Before you go through this article, make sure that you have gone through the previous article on **Distance and Displacement**.

We have discussed-

- Distance is the total length of the actual path travelled by a particle during its motion.
- Displacement is a vector drawn from the initial position to the final position of the particle.

In this article, we will discuss problems on distance and displacement.

A particle covers half of the circle of radius R. Then, the displacement and distance of the particle are respectively-

- 2πR, 0
- 2R, πR
- πR/2, 2R
- πR, R

Consider particle moves from point A to point B covering half of the circle as shown-

Magnitude of displacement of the particle

= Shortest distance between point A and point B

= Diameter of the circle

= 2R units

Distance covered by the particle

= Total length of the actual path

= Circumference of the circle / 2

= 2πR / 2

= πR units

Thus, Option (B) is correct.

A particle moves in a quarter circle of radius R. What are the values of distance and displacement respectively?

- πR/2, √2R
- πR, R
- 2√2R, πR/2
- πR/2, 0

Consider particle moves from point A to point B covering half of the circle as shown-

Distance covered by the particle

= Total length of the actual path

= Length of arc AB

= π/2 x R { Using the relation θ = arc / radius }

= πR/2 units

Magnitude of displacement of the particle

= Shortest distance between point A and point B

= Length of line AB

= √(OA^{2} + OB^{2})

= √(R^{2} + R^{2})

= √2R units

Thus, Option (A) is correct.

A mosquito flies from a point A(-1, 2, 3) m to a point B(2, -1, -2) m. Find the displacement of the mosquito.

We have-

- Initial position vector,
- Final position vector,

Displacement vector of the mosquito is given by-

So, we have-

Thus, displacement vector of the mosquito is-

The magnitude of displacement is given by-

||

= √(3^{2} + 3^{2} + 5^{2})

= √(9 + 9 + 25)

= √43 m

Thus, magnitude of displacement of the mosquito = **√43 m**.

A mosquito flies from corner A to the corner B of a cube of side length 5 m moving along its sides as shown. Find the displacement of the mosquito.

Consider the origin of Cartesian coordinate system to be present at point A.

Then-

- Coordinates of point A = (0, 0, 0)
- Coordinates of point B = (5, 5, 5)

Displacement vector of the mosquito is given by-

Thus, displacement vector of the mosquito is-

The magnitude of displacement is given by-

||

= √(5^{2} + 5^{2} + 5^{2})

= √(3 x 5^{2})

= 5√3 m

Thus, magnitude of displacement of the mosquito = **5****√3 m**.

A particle is moving in a circular path of radius R. The distance and displacement of the particle when it describes an angle of 60° from its initial position respectively are-

- πR/3, R
- πR/6, √2R
- πR/3, √2R
- πR/3, √2R

Consider particle moves from point A to point B covering an angle of 60º from its initial position as shown-

Distance covered by the particle

= Total length of the actual path

= Length of arc AB

= π/3 x R { Using the relation θ = arc / radius }

= πR/3 units

In Δ AOB,

- OA = OB = R
- ∠AOB = 60°

Now,

- We know, angles opposite to equal sides are equal. So, ∠OAB = ∠OBA.
- Also, we know sum of three angles in a triangle is 180°.
- So, we conclude ∠OAB = ∠OBA = 60°.

Since all the angles in ΔAOB are 60°, so ΔAOB is an equilateral triangle.

So, all the sides length must be same.

Thus, length of line AB = R units.

Now,

Magnitude of displacement of the particle

= Shortest distance between point A and point B

= Length of line AB

= R units

Thus, Option (A) is correct.

A player completes a circular path of radius R in 40 seconds. Its distance and displacement at the end of 2 minutes 20 seconds will be-

- 7πR, 2R
- 2R, 2R
- 2πR, 2R
- 7πR, R

We have number of revolutions made by the player in 40 seconds = 1.

So, number of revolutions made by the player in 2 minutes 20 seconds (140 seconds)

= x 140

= 3.5 revolutions

Distance covered in 3.5 revolutions

= 3.5 x Distance covered in 1 revolution

= 3.5 x Circumference of the circle

= 3.5 x 2πR

= 7πR units

Magnitude of displacement after 3.5 revolutions

= Diameter of the circle

= 2R units

Thus, Option (A) is correct.

Given that P is a point on a wheel rolling on horizontal road. The radius of the wheel is R. Initially, the point P is in contact with ground. The wheel rolls through half of the revolution. What is the displacement of point P?

Given-

- Initially, the point P is in contact with ground.
- The wheel rolls through half of the revolution.

After rolling through half revolution, point P is at the top most point of the wheel.

So, we have-

Here, d represents the displacement of point P.

Using Pythagoras theorem, we have-

d^{2} = (2R)^{2} + (πR)^{2}

d^{2} = 4R^{2} + π^{2}R^{2}

d^{2} = (4+π^{2})R^{2}

∴ d = R√(4+π^{2})

Thus, magnitude of displacement of point P = **R√(4+π ^{2}) m**.

To practice more problems on distance and displacement,

**Next Article-** **Speed and Velocity**

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Before you go through this article, make sure that you have gone through the previous article on **Motion in One Dimension**.

We have discussed-

- In one-dimensional motion, particle moves along a straight line.
- It is also known as rectilinear motion.

Various parameters related to motion are-

In this article, we will discuss about acceleration.

- The motion of a particle is said to be accelerated if its velocity changes with time.
- Example- A vehicle moving on a crowded road.
- It is a kind of non-uniform motion.

Rate of change of velocity of a particle is called as its acceleration. |

Mathematically,

The characteristics of acceleration are-

- It is a vector quantity.
- Acceleration can be positive, negative or zero.
- Acceleration is zero for a moving particle if it moves with constant velocity.
- The SI unit of acceleration is meter per second
^{2}(m/s^{2}). - The dimensional formula of acceleration is [M
^{0}L^{1}T^{-2}].

**Also Read-** **Speed and Velocity**

There are mainly following four types of acceleration-

- Uniform Acceleration
- Variable Acceleration
- Average Acceleration
- Instantaneous Acceleration

A particle is said to be moving with uniform acceleration if equal change in velocity takes place in equal intervals of time.
A particle is said to be moving with uniform acceleration if its velocity changes with a uniform rate. |

A particle is said to be moving with variable acceleration
if its velocity changes equally in unequal intervals of time or unequally in equal intervals of time. |

The average acceleration of a particle is that constant acceleration with which
the particle undergoes same change in velocity in a given time as it undergoes while moving with variable acceleration during the given time. |

Mathematically,

It is the ratio of total change in velocity of the particle to the total time interval in which this change in velocity takes place.

Consider-

- At any time t
_{1}, the velocity of the particle is v1. - At time t
_{2}, the velocity becomes v2.

Then for this interval, average acceleration is given by-

Instantaneous acceleration is the acceleration of particle at a particular instant of time. |

Mathematically,

Instantaneous acceleration is the limiting value of average acceleration as the time interval becomes infinitesimally small.

If Δv is the change in velocity of particle in the time interval Δt, then-

**Also Read-** **Distance & Displacement**

It is important to note the following points-

- Acceleration is always in the direction of increasing velocity.
- Change in velocity = Final velocity – Initial velocity

It is worth remembering the following relation-

We know, Average Velocity

= Total displacement / Total time taken

But if the motion is uniformly accelerated, then average velocity can also be written as-

- Negative acceleration is called as
**Retardation**or**Deceleration**. - It is responsible for decreasing the velocity of a particle.

To gain better understanding about Acceleration,

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Before you go through this article, make sure that you have gone through the previous article on **Motion in One Dimension**.

We have discussed-

- In one-dimensional motion, particle moves along a straight line.
- It is also known as rectilinear motion.

Various parameters related to motion are-

In this article, we will discuss about speed and velocity.

Rate of change of distance of a particle is called as its speed.
Distance covered by a particle per unit time is called as its speed. |

Mathematically,

Consider-

- A particle moves in a circular path of radius 7 m.
- It starts its journey from point A and ends at the same point in time 11 seconds.

Then, Speed of the particle

= Distance covered by the particle during its motion / Time taken

= Circumference of the circle / Time taken

= (2 x 22/7 x 7) / 11

= 4 m/s

The characteristics of speed are-

- It is a scalar quantity.
- Speed is always positive. It can never be negative.
- Speed can increase or decrease with time.
- Speed is never zero for a moving particle.
- The SI unit of speed is meter per second (m/s).
- The dimensional formula of speed is [M
^{0}L^{1}T^{-1}].

**Also Read-** **Distance Vs Displacement**

There are mainly following four types of speeds-

- Uniform Speed
- Variable Speed
- Average Speed
- Instantaneous Speed

A particle is said to be moving with uniform speed if it covers equal distances in equal intervals of time.
A particle is said to be moving with uniform speed if it moves continuously with constant speed. |

It is worth remembering-

The direction of motion of the particle may change while moving with constant speed.

A particle is said to be moving with variable speed
if it covers equal distances in unequal intervals of time or unequal distances in equal intervals of time. |

The average speed of a particle is that constant speed with which the particle covers the same distance
in a given time as it does while moving with variable speed during the given time. |

Mathematically,

It is the ratio of total distance travelled by the particle to the total time taken in which the distance is travelled.

If Δx is the distance travelled by particle in time Δt, then average speed is given by-

Instantaneous speed is the speed of particle at a particular instant of time. |

Mathematically,

Instantaneous speed is the limiting value of average speed as the time interval becomes infinitesimally small.

If Δx is the distance covered by a particle in the time interval Δt, then-

Rate of change of displacement of a particle is called as its velocity.
Rate of change of position vector of a particle is called as its velocity. |

Mathematically,

Its direction is same as that of displacement.

Consider-

- A particle moves in a circular path of radius 7 m.
- It starts its journey from point A and ends at the same point in time 11 seconds.

Velocity of the particle

= Displacement of the particle / Time taken

= 0 / 11

= 0 m/s

The characteristics of velocity are-

- It is a vector quantity.
- Velocity can be positive, negative or zero.
- Velocity can increase or decrease with time.
- Velocity may be zero for a moving particle.
- The SI unit of velocity is meter per second (m/s).
- The dimensional formula of velocity is [M
^{0}L^{1}T^{-1}].

There are mainly following four types of velocities-

- Uniform Velocity
- Variable Velocity
- Average Velocity
- Instantaneous Velocity

A particle is said to be moving with uniform velocity if it covers equal displacements in equal intervals of time.
A particle is said to be moving with uniform velocity if it moves continuously in the same direction with constant speed. |

It is worth remembering-

A particle is said to be moving with variable velocity
if it covers equal displacements in unequal intervals of time or unequal displacements in equal intervals of time.
A particle is said to be moving with variable velocity if either magnitude of velocity or direction or both changes during its motion. |

The average velocity of a particle is that constant velocity with which the particle undergoes same
displacement in a given time as it undergoes while moving with variable velocity during the given time. |

Mathematically,

It is the ratio of total displacement of the particle to the total time interval in which the displacement occurs.

Consider-

- At any time t
_{1}, the position vector of the particle is . - At time t
_{2}, the position vector is .

Then for this interval, average velocity is given by-

Instantaneous speed is the velocity of particle at a particular instant of time. |

Mathematically,

Instantaneous velocity is the limiting value of average velocity as the time interval becomes infinitesimally small.

If Δr is the displacement of particle in the time interval Δt, then-

Some important differences between speed and velocity are-

Speed |
Velocity |

Rate of change of distance of a particle is called as its speed. | Rate of change of displacement of a particle is called as its velocity. |

It is a scalar quantity. | It is a vector quantity. |

Speed is either positive or zero. It can never be negative. | Velocity can be positive, negative or zero. |

Speed is never zero for a moving particle. | Velocity can be zero for a moving particle. |

Speed tells nothing about the direction of motion of the particle. | Velocity tell the direction of motion of the body. |

It is important to note the following points-

According to convention,

- If the particle moves upwards or towards right, its velocity is taken positive.
- If the particle moves downwards or towards left, its velocity is taken negative.

Velocity tell the direction of motion of the particle as-

- If the velocity is positive, particle must be moving towards positive direction.
- If the velocity is negative, particle must be moving towards negative direction.

- If motion takes place in the same direction, then average speed and average velocity are same.
- This is because distance and displacement are then same.

- The magnitude of instantaneous velocity is always instantaneous speed.
- Instantaneous speed and instantaneous velocity differs only by direction.
- However, magnitude of average velocity is not always average speed.
- Speedometer of an automobile measures instantaneous speed of the automobile.

To gain better understanding about Speed and Velocity,

**Next Article-** **Acceleration and Retardation**

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Before you go through this article, make sure that you have gone through the previous article on **Motion in One Dimension**.

We have discussed-

- In one-dimensional motion, particle moves along a straight line.
- It is also known as rectilinear motion.

Various parameters related to motion are-

In this article, we will discuss about distance and displacement.

Distance is the total length of the actual path travelled by a particle during its motion. |

Consider-

- A particle moves in a circular path of radius R.
- It starts its journey from point A and ends at the same point.

Distance covered by the particle during its motion

= Circumference of the circle

= 2πR units

The characteristics of distance are-

- It depends upon the path followed by the particle.
- It is a scalar quantity.
- Distance is always positive. It can never be negative.
- Distance can not decrease with time.
- Distance is never zero for a moving particle.
- The SI unit of distance is meter (m).
- The dimensional formula of distance is [M
^{0}L^{1}T^{0}].

Displacement is a vector drawn from the initial position to the final position of the particle.
Its magnitude is equal to the shortest distance between the initial and final position. |

Mathematically, it is equal to the change in position vector of the particle.

Consider-

- A particle is initially at point A having position vector .
- The particle moves to point B having position vector .

Consider-

- A particle moves along a straight line from point O to point A.
- It then travels back from point A to point B as shown-

Magnitude of displacement of the particle

= Shortest distance between its initial and final position

= 70 m – 0 m

= 70 m

Direction of displacement is towards +ve X-axis.

It is important to note that here the distance covered by the particle = 100 m + 30 m = 130 m.

The characteristics of displacement are-

- It is independent of the path followed by the particle.
- It is a vector quantity.
- Displacement can be positive, negative or zero.
- Displacement can decrease with time.
- Displacement may be zero for a moving particle.
- The SI unit of displacement is meter (m).
- The dimensional formula of displacement is [M
^{0}L^{1}T^{0}].

Some important differences between distance and displacement are-

Distance |
Displacement |

It is the total length of the actual path travelled by a particle during its motion. | It is the shortest distance between the initial and final positions. |

It depends upon the path followed by the particle. | It is independent of the path followed by the particle. |

It is a scalar quantity. | It is a vector quantity. |

Distance is either positive or zero. It can never be negative. | Displacement can be positive, negative or zero. |

Distance can never decrease with time. | Displacement may decrease with time. |

Distance between a given set of initial and final position can have infinite values. | Displacement between a given set of initial and final position is unique. |

Distance is never zero for a moving particle. | Displacement may be zero for a moving particle. |

It is important to note the following points-

- Magnitude of displacement can never be greater than distance.
- Magnitude of displacement is equal to the distance if the particle moves along a straight line without changing the direction.

When a particle returns to its initial position,

- Its displacement is always zero.
- Its distance covered is not zero.

According to convention,

- If the particle moves upwards or towards right, its displacement is taken positive.
- If the particle moves downwards or towards left, its displacement is taken negative.

To gain better understanding about Distance and Displacement,

**Next Article-** **Problems On Distance & Displacement**

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A frame of reference is a frame in which an observer sits and makes observations. |

- A Cartesian Coordinate System is attached with the frame of reference.
- A clock is positioned in this system to measure time.
- An observer is always considered to be present at the origin O.

At any time, the position of a particle is described with the help of its three position coordinates (x, y, z).

Consider the particle is present at any point P in the space whose coordinates are (x, y, z).

The position vector of point P is a vector drawn from the origin O of the coordinate system to point P.

It is given by-

Frame of reference can be of two types-

- Inertial Frame of Reference
- Non-Inertial Frame of Reference

- A frame of reference which is either at rest or moving with constant velocity is known as inertial frame of reference.
- Newton’s laws are valid in this frame.

- A frame of reference moving with some acceleration is known as non-inertial frame of reference.
- Newton’s laws are not valid in this frame.

A body is said to be at rest if it does not change its position with time. A body is said to be in motion if it continually change its position with time. |

- Rest and motion are relative terms.
- When a particle moves, its position coordinates change with time.
- At one time, one or two or all three position coordinates may change.

Accordingly, we have following three types of motion-

- Motion in one dimension
- Motion in two dimensions
- Motion in three dimensions

The motion of a particle is said to be in one dimension if
only one out of the three coordinates specifying the position of the particle change with time. |

- In such a motion, the particle moves along a straight line.
- One dimensional motion is sometimes known as
**rectilinear**or**linear motion**.

Examples of motion in 1 dimension are-

- A car moving on a straight road
- A ball thrown vertically up
- A stone dropped into a well
- A ball dropped from a certain height above the ground

The motion of a particle is said to be in two dimensions if
any two out of the three coordinates specifying the position of the particle change with time. |

In such a motion, the particle moves in a plane.

Examples of motion in two dimensions are-

- Projectile motion
- Circular motion
- A carom coin rebounding smoothly from the side of the board
- Motion of a boat in a river
- Motion of an insect on a floor

The motion of a particle is said to be in three dimensions if
all the three coordinates specifying the position of the particle change with time. |

In such a motion, the particle moves in a space.

Examples of motion in three dimensions are-

- Random motion of a gas molecule
- A kite flying in the sky
- A bird flying in the sky

We discuss the following parameters in detail while studying the motion of any particle-

- Distance
- Displacement
- Speed
- Velocity
- Acceleration

“Rest and Motion are relative terms.” Justify this statement.

- It is rightly mentioned that the rest and motion are relative terms.
- This is because they depend on the observer’s frame of reference.
- A body at rest with respect to one observer might be in motion with respect to another observer.
- A body in motion with respect to one observer might be at rest with respect to another observer.

Consider the following examples-

- A person sitting in a moving train is at rest with respect to fellow passengers.
- While person sitting in a moving train is in motion with respect to a person standing on the platform.

Consider the following scenario-

- Two persons A and B are standing in a truck moving with velocity v.
- Person C is standing on the ground.

Here,

- A and B are at rest with respect to each other.
- A and B moves with velocity V in the positive X direction with respect to C.
- C moves with velocity V in the negative X direction with respect to A and B.

Is it possible to achieve state of absolute rest or absolute motion?

- A body at rest with respect to all other bodies is called in state of absolute rest.
- A body moving with respect to all other bodies is called in state of absolute motion.

Talking about absolute rest,

- It is impossible to achieve the state of absolute rest.
- This is because all heavenly bodies are moving with respect to each other.

Talking about absolute motion,

- It is impossible to achieve the state of absolute motion.
- This is because there exists no reference point which is absolutely fixed in the space.

To gain better understanding about Rest and Motion,

**Next Article-** **Distance and Displacement**

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Before you go through this article, make sure that you have gone through the previous article on **Electric Lines of Force**.

We have discussed-

- Electric lines of force represent the electric field around the charge.
- Electric lines of force are the imaginary curves.
- But the field they represent is real.

In this article, we will discuss about Electric Flux.

The total number of electric lines of force passing perpendicularly through a given area
held inside an electric field is called as electric flux. |

Electric flux is denoted by the symbol **Φ _{E}**.

The electric flux passing through any given area depends upon the following factors-

**Also Read-** **Electric Lines of Force**

Electric flux ∝ Strength of Electric Field

This is because-

- Stronger the electric field, closer are the electric lines of force.
- Closer the electric lines of force, greater the amount of electric flux passing through the surface.

Electric flux ∝ Area of the Surface

This is because-

- Greater the area of the surface, greater the amount of electric lines of force passing through it.
- Therefor, greater is the electric flux passing through the area.

Electric flux ∝ cosθ

This is because-

- As the angle (θ) between electric field and area vector increases, the number of electric lines of force passing through the area decreases.

On combining these factors, we get the following electric flux formula.

**Also Read-** **Electric Field**

Electric flux Φ_{E} passing through any given area held inside an electric field of strength is given by-

- The SI unit of electric flux is NC
^{-1}m^{2}. - Electric flux is a scalar quantity.

Electric flux passing per unit area through the surface is called as electric flux density.

Mathematically,

- The SI unit of electric flux density is NC
^{-1}. - Electric flux density is a scalar quantity.

It is worth mentioning the following notes-

- Area vector for a 2D surface is always perpendicular to its surface.
- Area vector for a 3D surface is always outward normal to its surface.

- The formula Φ
_{E}= = EAcosθ is valid if and only if the value of electric field E at every point lying on the area A is same.

- The formula Φ
_{E}= ∫ is valid everywhere. - When electric field is uniform, comes out of the integration and the formula reduces to Φ
_{E}= .

Calculate the electric flux passing through the surface kept in the electric field with its plane perpendicular to the electric field as shown-

We have-

Given-

- Electric field strength (E) = 100 N/C
- Area of the surface (A) = 2 m x 2 m = 4 m
^{2} - Angle between and = 0°

We know, electric flux Φ_{E} passing through any surface is given by-

Φ_{E} = EAcosθ

Substituting the values, we get-

Φ_{E} = 100 x 4 x cos 0°

Φ_{E} = 100 x 4 x 1

Φ_{E} = 400 NC^{-1}m^{2}

Thus, electric flux passing through the given surface = 400 NC^{-1}m^{2}.

Calculate the electric flux passing through the surface kept in the electric field with its plane parallel to the electric field as shown-

We have-

Given-

- Electric field strength (E) = 100 N/C
- Area of the surface (A) = 2 m x 2 m = 4 m
^{2} - Angle between and = 90°

We know, electric flux Φ_{E} passing through any surface is given by-

Φ_{E} = EAcosθ

Substituting the values, we get-

Φ_{E} = 100 x 4 x cos 90°

Φ_{E} = 100 x 4 x 0

Φ_{E} = 0

Thus, no electric flux passes through the given surface.

Calculate the electric flux passing through the circular disk of radius 7 m placed in the electric field as shown-

We have-

Given-

- Electric field strength (E) = 100 N/C
- Radius of circular disk (R) = 7 m
- Angle between and = 60°

Now,

Area of the disk surface (A)

= πR^{2}

= (22/7) x 7 x 7

= 154 m^{2}

We know, electric flux Φ_{E} passing through any surface is given by-

Φ_{E} = EAcosθ

Substituting the values, we get-

Φ_{E} = 100 x 154 x cos 60°

Φ_{E} = 100 x 154 x 0.5

Φ_{E} = 7700 NC^{-1}m^{2}

Thus, electric flux passing through the given circular disk = 7700 NC^{-1}m^{2}.

If = , calculate the electric flux through a surface of area 20 units in Y-Z plane.

Given-

- Electric field () =
- Area of the surface () =

We know, electric flux Φ_{E} passing through any surface is given by-

Φ_{E} =

Substituting the values, we get-

Φ_{E} = ().

Φ_{E} = 120 units

Thus, electric flux passing through the given surface = 120 units.

A cylinder of radius R is placed in a uniform electric field E with its axis parallel to the field as shown-

Calculate the electric flux passing through-

- Left plane surface
- Right plane surface
- Curved surface
- Complete cylinder

Given-

- Electric field strength = E
- Radius of the cylinder = R

We know, electric flux Φ_{E} passing through any surface is given by-

Φ_{E} = EAcosθ

Electric flux through left plane surface (Φ_{1})

= E x πR^{2} x cos 180°

= -EπR^{2}

Thus, electric flux passing through the left plane surface = -EπR^{2}.

Electric flux through right plane surface (Φ_{2})

= E x πR^{2} x cos 0°

= EπR^{2}

Thus, electric flux passing through the right plane surface = EπR^{2}.

Electric flux through curved surface (Φ_{3})

= E x 2πRL x cos 90°

= 0

Thus, electric flux passing through the curved surface = 0.

Electric flux through complete cylinder (Φ_{E})

= Φ_{1} + Φ_{2} + Φ_{3}

= -EπR^{2} + EπR^{2} + 0

= 0

Thus, electric flux passing through the complete cylinder = 0.

A positive charge of 17.7 μC is placed at the center of a hollow sphere of radius 0.5 m. Calculate the flux density through the surface of the sphere.

Given-

- Radius of hollow sphere (R) = 0.5 m
- Charge at center (Q) = 17.7 μC

Electric field at the surface of a hollow sphere due to the charge kept at the center is constant.

The electric field at the surface is directed outward normal.

It is given by-

Substituting the values, we get-

Electric field at surface

= (9 x 10^{9} x 17.7 x 10^{-6}) / (0.5)^{2}

= 637.2 x 10^{3}

= 6.37 x 10^{5} N/C

We have-

Electric flux passing through surface of the sphere

= EAcosθ

= E x 4πR^{2} x cosθ

= 6.37 x 10^{5} x (4 x 3.14 x 0.5 x 0.5) x cos0°

= 20 x 10^{5} NC^{-1}m^{2}

Electric flux density

= Total electric flux passing through the surface / Area of the surface

= 20 x 10^{5} / (4 x 3.14 x 0.5 x 0.5)

= 20 x 10^{5} / 3.14

= 6.36 x 10^{5}

= 6.4 x 10^{5} N/C

Thus, electric flux density through the surface = 6.4 x 10^{5} N/C.

To gain better understanding about Electric Flux,

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Before you go through this article, make sure that you have gone through the previous article on **Electric Field**.

We have discussed-

- Electric field is the space around an electric charge in which its effect can be experienced.
- Every charge creates an electric field in the space around itself.
- This electric field extends till infinity.

There are following 4 types of electric field-

In this article, we will discuss about Electric Lines of Force.

An electric field line or **Electric Line of Force** may be defined as-

It is defined as imaginary line in an electric field, the tangent to which at any point on it represents the direction of electric field at that point.
It is defined as the imaginary path traversed by a unit positive charge in an electric field.
It is defined as the curve along which a unit positive charge tends to move if free to do so in an electric field. |

It is important to note that-

- The electric lines of force are imaginary curves.
- But the electric field represented by them is real.
- The term “lines of force” is misnomer.
- It is more appropriate to call these lines as
**“electric field lines”**.

Following are some of the important properties of electric lines of force-

- The lines of force are continuous smooth curves.
- They do not have any breaks.

- The lines of force originate out of a positive charge and terminates on a negative charge.
- If there exists a single charge, then lines of force end at infinity.
- The lines of force cannot form closed loops.

- The tangent to a line of force at any point gives the direction of electric field at that point.

- No two lines of force can cross each other.

- The lines of force have a tendency to contract longitudinally.
- The lines of force have a tendency to exert a lateral pressure on neighboring lines of force.

- The lines of force are always normal to the surface of a conductor on which the charges are in equilibrium.

- The relative closeness of the lines of force gives a measure of the strength of the electric field.
- Closer the lines of force, stronger is the electric field.

- The lines of force do not pass through a conductor.

Let us represent the electric lines of force for the following-

- Electric lines of force of a positive point charge
- Electric lines of force of a negative point charge
- Electric lines of force of an electric dipole
- Electric lines of force for a system of two equal positive point charges

If there exists only one positive point charge, then-

- The electric lines of force originate from that charge.
- The electric lines of force terminate at infinity.

If there exists only one negative point charge, then-

- The electric lines of force originate from infinity.
- The electric lines of force terminate at that charge.

Electric dipole is a system of two equal and opposite charges separated by a small distance.

If there exists an electric dipole, then-

- The electric lines of force originate from the positive charge.
- The electric lines of force terminate at the negative charge.

If there exists a system of two positive charges, then-

- They exert a lateral pressure on account of repulsion between like charges.
- The electric field is zero at the middle point on the line joining the two charges.
- This point is called neutral point from which no line of force passes.

Why electric lines of force are smooth continuous curves?

Electric lines of force are smooth continuous curves without any breaks because-

- Their discontinuity would indicate that the charge suddenly vanishes and reappears.
- This is absurd.

Why no two electric lines of force can intersect each other?

No two electric lines of force can intersect each other because-

- If they intersect, then there will be two tangents at the point of intersection.
- Hence, two directions of the electric field at the same point will exist which is impossible.

Why electric lines of force contract longitudinally and exerts a lateral pressure?

- Electric lines of force contract length-wise on account of attraction between unlike charges.
- Electric lines of force exert a lateral pressure on account of repulsion between like charges.

Why electric lines of force are always normal to the surface of a conductor on which the charges are in equilibrium?

This is so because if the lines of force are not normal to the surface of a conductor, then-

- There exists a component of the field parallel to the surface.
- This would cause the electrons to move and set up a current on the surface.
- But no current flows in the equilibrium condition.

Why electric lines of force never pass through a conductor?

Electric lines of force never pass through a conductor because electric field inside a charged conductor is zero.

To gain better understanding about Electric Lines of Force,

**Next Article-** **Electric Flux**

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Before you go through this article, make sure that you have gone through the previous article on **Electric Field**.

We have discussed-

- Electric field is the space around an electric charge in which its effect can be experienced.
- Electric field intensity at any point is the strength of electric field at that point.

There are following 4 types of electric fields-

It is important to remember the following formulas while solving problems based on electric field-

Electric field intensity (E) at any point is the force (F) experienced by a unit positive charge placed at that point.

The direction of electric field is same as the direction of force.

Force experienced by charge q when placed in an electric field having electric field intensity E is given by-

The direction of force is same as the direction of electric field.

Acceleration of an electric charge q_{0} having mass m in an electric field having electric field intensity is given by-

Electric field intensity due to a point charge Q at distance ‘r’ units from it is given by-

Electric field intensity at any point P due to the system of N charges is given by-

** = + + + ……. + **

In this article, we will discuss practice problems based on Electric Field.

A particle having a charge 1 x 10^{-9} C experiences a downward force of 1.5 x 10^{-6} N when placed in an electric field. Calculate the electric field intensity.

Given-

- Charge (q) = 1 x 10
^{-9}C - Force (F) = 1.5 x 10
^{-6}N

We know,

Electric field intensity (E) is given by-

Substituting the given values, we get-

Electric field intensity (E)

= (1.5 x 10^{-6} N) / (1 x 10^{-9} C)

= 1.5 x 10^{3} N/C

Thus, electric field intensity is 1.5 x 10^{3} N/C vertically downward.

Calculate the electric field strength required to just support a water drop of mass 10^{-3} kg and having a charge 1.6 x 10^{-19} C.

Given-

- Mass of water drop (m) = 10
^{-3}kg - Charge (q) = 1.6 x 10
^{-19}C

Let electric field intensity required to just support the water drop = E.

To just support the water drop, the condition is-

Force on water drop due to electric field = Weight of water drop

q x E = m x g

Substituting the given values, we get-

1.6 x 10^{-19} C x E = 10^{-3} kg x 9.8 m/s^{2}

E = (9.8 x 10^{-3} kg m/s^{2}) / 1.6 x 10^{-19} C

E = 6.125 x 10^{16} N/C

Thus, electric field strength required to just support the water drop is 6.125 x 10^{16} N/C.

An electron enters with a velocity of 5 x 10^{6} m/s along the positive direction of an electric field of intensity 10^{3} N/C. Calculate the time taken by the electron to come temporarily to rest. Mass of the electron is 9.11 x 10^{-31} kg.

Given-

- Initial velocity of the electron (u) = 5 x 10
^{6}m/s - Electric Field Intensity (E) = 10
^{3}N/C - Mass of the electron (m) = 9.11 x 10
^{-31}kg - Final velocity of the electron (v) = 0

We know-

Acceleration of an electric charge q_{0} having mass m in an electric field having electric field intensity is given by-

Substituting the given values, we get-

Acceleration of the electron (a)

= (1.6 x 10^{-19} C x 10^{3} N/C) / (9.11 x 10^{-31} kg)

= 0.175 x 10^{15} m/s^{2}

This is actually deceleration which causes the electron to slow down and makes it stop momentarily.

By Newton’s first equation of motion, we know v = u + at.

Substituting the given values, we get-

0 = 5 x 10^{6} m/s – 0.175 x 10^{15} m/s^{2} x t

0.175 x 10^{15} m/s^{2} x t = 5 x 10^{6} m/s

t = (5 x 10^{6} m/s) / (0.175 x 10^{15} m/s^{2})

t = 28.57 x 10^{-9} s

t = 2.9 x 10^{-8} s

Thus, time taken by the electron to come temporarily to rest is 2.9 x 10^{-8} seconds.

An electron moves a distance of 6 cm when accelerated from rest by an electric field of strength 2 x 10^{4} N/C. Calculate the time of travel. The mass and charge of electron are 9 x 10^{-31} kg and 1.6 x 10^{-19} C respectively.

Given-

- Distance traveled by the electron (s) = 6 cm = 0.06 m
- Initial velocity of the electron (u) = 0 m/s
- Electric Field Intensity (E) = 2 x 10
^{4}N/C - Mass of the electron (m) = 9 x 10
^{-31}kg - Charge on the electron (q) = 1.6 x 10
^{-19}C

We know-

Acceleration of an electric charge q_{0} having mass m in an electric field having electric field strength is given by-

Substituting the given values, we get-

Acceleration of the electron (a)

= (1.6 x 10^{-19} C x 2 x 10^{4} N/C) / (9 x 10^{-31} kg)

= 0.36 x 10^{16} m/s^{2}

By Newton’s second equation of motion, we know s = ut + ( )at^{2}.

Substituting the given values, we get-

0.06 m = 0 x t + ( ) x 0.36 x 10^{16} m/s^{2} x t^{2}

0.12 m = 0.36 x 10^{16} m/s^{2} x t^{2}

0.12 s^{2} = 0.36 x 10^{16} x t^{2}

t^{2} = 0.12 / (0.36 x 10^{16}) s^{2}

t^{2} = 0.33 x 10^{-16} s^{2}

t = 0.57 x 10^{-8} s

t = 5.7 x 10^{-9} s

t = 5.7 ns

Thus, required time of travel is 5.7 nanoseconds.

Two point charges of 5 x 10^{-19} C and 20 x 10^{-19} C are separated by a distance of 2 m. Find the point on the line joining them at which electric field intensity is zero.

Given-

- Charge (q
_{1}) = 5 x 10^{-19}C - Charge (q
_{2}) = 20 x 10^{-19}C - Distance between the two charges = 2 m.

- Electric field due to charge q
_{1}is towards right along the line joining the two charges. - Electric field due to charge q
_{2}is towards left along the line joining the two charges.

Let-

- Electric field intensity is zero at point P on the line joining the two charges.
- Electric field intensity at point P due to charge q
_{1}is E_{1}. - Electric field intensity at point P due to charge q
_{2}is E_{2}. - Distance of point P from charge q
_{1}is x m. - Distance of point P from charge q
_{2}is (2-x) m.

For electric field to be zero at point P,

Electric field intensity at point P due to q_{1} = Electric field intensity at point P due to q_{2}

Substituting the given values, we get-

E_{1} = E_{2}

K x ( ) = K x { }

q1 x (2 – x)^{2} = q_{2} x x^{2}

5 x 10^{-19} C x (2 – x)^{2} = 20 x 10^{-19} C x x^{2}

(2 – x)^{2} = 4x^{2}

4 + x^{2} – 4x = 4x^{2}

3x^{2} + 4x – 4 = 0

On solving, we get x = 0.67 m.

Thus, electric field is zero at 0.67 meters to the right of charge q_{1}.

Two point charges q_{1} and q_{2} of 10^{-8} C and -10^{-8} C respectively are placed 0.1 m apart. Calculate the electric fields at points A, B and C as shown in figure-

Let-

- Electric field intensity at point A due to charge q
_{1}= E_{1A} - Electric field intensity at point A due to charge q
_{2}= E_{2A}

It is clear that-

- Electric field at point A due to charge q
_{1}is directed towards right. - Electric field at point A due to charge q
_{2}is directed towards right.

Thus, Net electric field intensity at point A

= |E_{1A}| + |E_{2A}|

= K x ( ) + K x ( )

= ( ) x (q_{1} + q_{2})

= (K x 400) x (10^{-8} C + 10^{-8} C)

= K x 400 x (2 x 10^{-8} C)

= K x 800 x 10^{-8} C

= 9 x 10^{9} x 800 x 10^{-8}

= 72000

= 7.2 x 10^{4} N/C

Thus, electric field intensity at point A is 7.2 x 10^{4} N/C directed towards right.

Let-

- Electric field intensity at point B due to charge q
_{1}= E_{1B} - Electric field intensity at point B due to charge q
_{2}= E_{2B}

It is clear that-

- Electric field at point B due to charge q
_{1}is directed towards left. - Electric field at point B due to charge q
_{2}is directed towards right.

Thus, Net electric field intensity at point B

= |E_{1B}| – |E_{2B}|

= K x ( ) – K x ( )

= K x { – }

= K x 10^{-8} x (400 – 44.44)

= 9 x 10^{9} x 10^{-8} x 355.56

= 32000

= 3.2 x 10^{4} N/C

Thus, electric field intensity at point A is 7.2 x 10^{4} N/C directed towards left.

Let-

- Electric field intensity at point C due to charge q
_{1}= E_{1C} - Electric field intensity at point C due to charge q
_{2}= E_{2C}

It is clear that-

- Electric field at point C due to charge q
_{1}is directed away from q_{1}along the line joining q_{1}and C. - Electric field at point C due to charge q
_{2}is directed towards q_{2}along the line joining q_{2}and C. - Angle between and is 120°.

Now,

Electric field intensity at point C due to charge q_{1}

= K x ( )

= K x 10^{-8} x 100

= 9 x 10^{9} x 10^{-8} x 100

= 9000 N/C directed away from q_{1} along the line joining q_{1} and point C.

Similarly,

Electric field intensity at point C due to charge q_{2}

= 9000 N/C directed towards q_{2} along the line joining q_{2} and point C.

Net electric field at point C

= sqrt { E^{2}_{1C} + E^{2}_{2C} + 2 x E_{1C} x E_{2C} x cos120° }

= sqrt { (9000)^{2} + (9000)^{2} + 2 x 9000 x 9000 x ( ) }

= sqrt { 2 x 81 x 10^{6} – 81 x 10^{6} }

= sqrt { 81 x 10^{6} }

= 9 x 10^{3} N/C

Thus, electric field intensity at point C is 9 x 10^{3} N/C.

The direction of electric field is along the bisector of the angle between and towards right.

This is because and are equal in magnitude.

To practice more problems based on Electric Field & Electric Field Intensity,

**Next Article-** **Electric Field Lines**

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Electric Field is defined as-

The space around an electric charge in which its effect can be experienced is called as its electric field.
It is the space around an electric charge in which any other charge experiences the coulomb force. |

- The charge which produces the electric field is called as a source charge.
- The charge which tests the effect of source charge is called as a test charge.

There are following 4 types of electric field-

- Uniform Electric Field
- Non-Uniform Electric Field
- Variable Electric Field
- Constant Electric Field

It is an electric field at every point of which a unit positive charge experiences the same electric force.

**OR**

It is an electric field at every point of which the intensity of electric field remains constant in magnitude and direction.

In uniform electric field, the electric lines of force are parallel and equidistant.

It is an electric field at different points of which a unit positive charge experiences the different electric force.

**OR**

It is an electric field at different points of which the intensity of electric field is different in magnitude and direction.

In non-uniform electric field, the electric lines of force are non-parallel and non-equidistant.

The electric field which changes with respect to time is called as variable electric field.

E = f(t)

The electric field which does not depend upon time is called as constant or non-variable electric field.

E ≠ f(t)

Electric Field Intensity is defined as-

Electric field intensity at any point is the strength of electric field at that point.
Electric field intensity at any point is the force experienced by a unit positive charge placed at that point. |

If is the force acting on a small test charge q_{0} at any point in the electric field, then electric field intensity at that point is given by-

It is important to note that-

- The test charge q
_{0}may have its own electric field. - This may modify the electric field of the source charge.
- So to minimize this effect, we write as-

From the above relation,

Force experienced by any point charge q_{0} in the electric field of intensity may be calculated as-

= q_{0} .

We know F = ma.

So, acceleration of electric charge q_{0} having mass m in electric field may be given as-

Consider-

- A point charge Q is placed at any point O.
- Electric field intensity due to charge Q has to be calculated at any point P.
- The distance between point O and point P is r units.

To calculate the electric field intensity at point P, we place a point charge q_{0} at point P.

According to Coulomb’s law,

Force at point P is given by-

Also, we know = / q_{0}

So,

This formula gives the electric field intensity due to a point charge Q at distance ‘r’ units from it.

This formula shows-

- Electric field intensity due to any point charge is inversely proportional to the square of the distance from it.
- As the distance from charge increases, its electric field intensity decreases.
- Electric field intensity becomes zero at infinity.

According to principle of superposition of electric charges,

Electric field intensity at a point due to a system of charges is the vector sum of the field intensities due to individual point charges.

Hence, the electric field intensity at any point P due to the system of N charges is-

** = + + + ……. + **

- Every charge creates an electric field in the space around itself.
- This electric field extends till infinity.
- Electric field due to a point charge at its own location is undefined.

- Electric field is a vector quantity since it possess both magnitude and direction.
- The electric field due to a positive point charge is radially outwards away from the charge.
- The electric field due to a negative point charge is radially inwards towards the charge.

- Electric field intensity is a vector quantity since it possess both magnitude and direction.
- The direction of is same as that of .
- The SI unit of electric field intensity is N/C (Newton per Coulomb).
- The dimensions of electric field intensity are [MLT
^{-3}A^{-1}].

The value of electric field intensity due to point charge at any point P depends on-

- Magnitude of the charge
- Distance of point P from the charge

It does not depend upon the test charge.

- Positive charges like alpha, proton, deutron etc experiences forces in the direction of the electric field.
- Negative charges like electron experiences force in the direction opposite to the electric field.
- Neutron experiences no force.

A point charge q_{0} is kept in an electric field of strength E and experiences a force F, then-

- E < F / q
_{0}if field is diverging - E = F / q
_{0}if field is uniform - E > F / q
_{0}if field is converging

To gain better understanding about Electric Field & Electric Field Intensity,

**Next Article-** **Problems On Electric Field**

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Before you go through this article, make sure that you have gone through the previous article on **Photoelectric Effect**.

We have discussed-

- Photoelectric Effect is a phenomenon of emission of electrons from metal surfaces.
- Photoelectric Effect occurs when a light of suitable frequency falls on the metal surface.

In this article, we will discuss about Photoelectric Cell.

Photoelectric Cell is a device that converts light energy into electrical energy. |

Also, it may also be defined as-

- Photoelectric cell is an electron tube or a photo tube.
- Electrons initiating an electric current in this electron tube originate by photoelectric emission.

Let us discuss the construction and working of a photoelectric cell.

- Photoelectric cell consists of an evacuated glass containing anode and cathode.
- The cathode is semi cylindrical.
- The cathode is coated with a photo sensitive alkali metal like rubidium, molybdenum etc.
- This is to reduce the work function.

- When light of suitable frequency is incident on the cathode, it emits the photoelectrons.
- These photoelectrons are attracted by the anode which is positively charged.
- This flow of electrons result in a flow of current.
- The flowing current is proportional to the intensity of incident radiation.
- This current may be further increased by filling some inert gas like argon into the glass.
- The photoelectrons emitted by cathode ionize the gas by collision.
- This collision results in increasing the current.
- The positive ions move towards the cathode.
- Negative ions and electrons move towards the anode.

Obviously, there is no conduction in the absence of light in the phototube.

Photoelectric cell has its applications in the following-

- Burglar alarm
- Counting machines
- Street lights
- Home security system
- Fire alarm

- A burglar alarm is fitted near the lock of almirah.
- A thief comes and makes the light fall on the lock by match stick, torch etc.
- The light is received by the photoelectric cell.
- The flow of electrons start from cathode to anode.
- This completes the circuit and alarm starts ringing.

- Counting machines may be operated by using a photoelectric cell.
- Whenever a currency note obstructs the incident light, flow of current stops for that duration.
- This advances the machine by count one.

- Street lights may be operated by using a photoelectric cell.
- The switch of the relay is attached with the photoelectric cell.
- When the intensity of sunlight decreases below the required value, the switch is activated.
- This switch on the street lights automatically.

- In a home security system, the light is produced by a source at one end of the circuit.
- This light falls on a photoelectric cell sensor located at some distance away from it.
- The electrons are emitted by the cathode.
- When somebody walks between the light source and the sensor, the path of the beam is blocked.
- The sensor registers a drop in light levels and sends a signal to the control box.
- The control box is connected to either siren or a telephone autodialer.
- The control box on receiving the signal responds accordingly.

- A photoelectric cell with a required circuit is installed in sensitive places.
- When the fire breaks out, light falls on the photoelectric cell.
- This causes the photoelectric emission which causes the current flow.
- This makes the bell connected to the circuit start to ring.

To gain better understanding about Photoelectric Cell,

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