Tag: Electric Charge and Field Objective Question

Electric Field Lines | Properties | Examples

Electric Field-

 

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.

 

Electric Field Lines-

 

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.

OR

It is defined as the imaginary path traversed by a unit positive charge in an electric field.

OR

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”.

 

Properties of Electric Field Lines-

 

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

 

Property-01:

 

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

 

Property-02:

 

  • 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.

 

Property-03:

 

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

 

Property-04:

 

  • No two lines of force can cross each other.

 

Property-05:

 

  • 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.

 

Property-06:

 

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

 

Property-07:

 

  • 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.

 

Property-08:

 

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

 

Electric Lines Of Force For Different Charged Conductors-

 

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

 

Case-01: For Positive Point Charge-

 

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.

 

 

Case-02: For Negative Point Charge-

 

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.

 

 

Case-03: For Electric Dipole-

 

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.

 

 

Case-04: For A System Of Two Equal Positive Point Charges-

 

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.

 

 

Conceptual Questions Based On Electric Lines Of Force-

 

Question-01:

 

Why electric lines of force are smooth continuous curves?

 

Answer-

 

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.

 

Question-02:

 

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

 

Answer-

 

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.

 

 

Question-03:

 

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

 

Answer-

 

  • 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.

 

Question-04:

 

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

 

Answer-

 

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 \overrightarrow{E} 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.

 

Question-05:

 

Why electric lines of force never pass through a conductor?

 

Answer-

 

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,

Watch this Video Lecture

 

Next Article- Electric Flux

 

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Electric Field Problems | Important Formulas

Electric Field-

 

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-

 

 

Important Formulas-

 

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

 

Formula-01:

 

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.

 

Formula-02:

 

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.

 

Formula-03:

 

Acceleration of an electric charge q0 having mass m in an electric field having electric field intensity \overrightarrow{E} is given by-

 

 

Formula-04:

 

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

 

 

Formula-05:

 

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

\overrightarrow{E} = \overrightarrow{{E}_1} + \overrightarrow{{E}_2} + \overrightarrow{{E}_3} + ……. + \overrightarrow{E{_N}}

 

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

 

PRACTICE PROBLEMS BASED ON ELECTRIC FIELD-

 

Problem-01:

 

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.

 

Solution-

 

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 103 N/C

 

Thus, electric field intensity is 1.5 x 103 N/C vertically downward.

 

Problem-02:

 

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.

 

Solution-

 

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/s2

E = (9.8 x 10-3 kg m/s2) / 1.6 x 10-19 C

E = 6.125 x 1016 N/C

 

Thus, electric field strength required to just support the water drop is 6.125 x 1016 N/C.

 

Problem-03:

 

An electron enters with a velocity of 5 x 106 m/s along the positive direction of an electric field of intensity 103 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.

 

Solution-

 

Given-

  • Initial velocity of the electron (u) = 5 x 106 m/s
  • Electric Field Intensity (E) = 103 N/C
  • Mass of the electron (m) = 9.11 x 10-31 kg
  • Final velocity of the electron (v) = 0

 

Step-01: Calculating Acceleration Of Electron-

 

We know-

Acceleration of an electric charge q0 having mass m in an electric field having electric field intensity \overrightarrow{E} is given by-

 

 

Substituting the given values, we get-

Acceleration of the electron (a)

= (1.6 x 10-19 C x 103 N/C) / (9.11 x 10-31 kg)

= 0.175 x 1015 m/s2

 

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

 

Step-02: Calculating Time Required To Come To Rest-

 

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

 

Substituting the given values, we get-

0 = 5 x 106 m/s – 0.175 x 1015 m/s2 x t

0.175 x 1015 m/s2 x t = 5 x 106 m/s

t = (5 x 106 m/s) / (0.175 x 1015 m/s2)

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.

 

Problem-04:

 

An electron moves a distance of 6 cm when accelerated from rest by an electric field of strength 2 x 104 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.

 

Solution-

 

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 104 N/C
  • Mass of the electron (m) = 9 x 10-31 kg
  • Charge on the electron (q) = 1.6 x 10-19 C

 

Step-01: Calculating Acceleration Of Electron-

 

We know-

Acceleration of an electric charge q0 having mass m in an electric field having electric field strength \overrightarrow{E} is given by-

 

 

Substituting the given values, we get-

Acceleration of the electron (a)

= (1.6 x 10-19 C x 2 x 104 N/C) / (9 x 10-31 kg)

= 0.36 x 1016 m/s2

 

Step-02: Calculating Time Of Travel-

 

By Newton’s second equation of motion, we know s = ut + ( \frac{1}{2} )at2.

 

Substituting the given values, we get-

0.06 m = 0 x t + ( \frac{1}{2} ) x 0.36 x 1016 m/s2 x t2

0.12 m = 0.36 x 1016 m/s2 x t2

0.12 s2 = 0.36 x 1016 x t2

t2 = 0.12 / (0.36 x 1016) s2

t2 = 0.33 x 10-16 s2

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.

 

Problem-05:

 

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.

 

Solution-

 

Given-

  • Charge (q1) = 5 x 10-19 C
  • Charge (q2) = 20 x 10-19 C
  • Distance between the two charges = 2 m.

 

 

  • Electric field due to charge q1 is towards right along the line joining the two charges.
  • Electric field due to charge q2 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 q1 is E1.
  • Electric field intensity at point P due to charge q2 is E2.
  • Distance of point P from charge q1 is x m.
  • Distance of point P from charge q2 is (2-x) m.

 

 

For electric field to be zero at point P,

Electric field intensity at point P due to q1 = Electric field intensity at point P due to q2

 

Substituting the given values, we get-

E1 = E2

K x ( \frac{q_{1}}{x^{2}} ) = K x { \frac{q2}{(2-x))^{2}} }

q1 x (2 – x)2 = q2 x x2

5 x 10-19 C x (2 – x)2 = 20 x 10-19 C x x2

(2 – x)2 = 4x2

4 + x2 – 4x = 4x2

3x2 + 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 q1.

 

Problem-06:

 

Two point charges q1 and q2 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-

 

 

Solution-

 

Calculating Electric Field Intensity At Point A-

 

Let-

  • Electric field intensity at point A due to charge q1 = E1A
  • Electric field intensity at point A due to charge q2 = E2A

 

It is clear that-

  • Electric field at point A due to charge q1 is directed towards right.
  • Electric field at point A due to charge q2 is directed towards right.

 

Thus, Net electric field intensity at point A

= |E1A| + |E2A|

= K x ( \frac{q_{1}}{0.05^{2}} ) + K x ( \frac{q_{2}}{0.05^{2}} )

= ( \frac{K}{0.0025} ) x (q1 + q2)

= (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 109 x 800 x 10-8

= 72000

= 7.2 x 104 N/C

 

Thus, electric field intensity at point A is 7.2 x 104 N/C directed towards right.

 

Calculating Electric Field Intensity At Point B-

 

Let-

  • Electric field intensity at point B due to charge q1 = E1B
  • Electric field intensity at point B due to charge q2 = E2B

 

It is clear that-

  • Electric field at point B due to charge q1 is directed towards left.
  • Electric field at point B due to charge q2 is directed towards right.

 

Thus, Net electric field intensity at point B

= |E1B| – |E2B|

= K x ( \frac{q_{1}}{0.05^{2}} ) – K x ( \frac{q_{2}}{0.15^{2}} )

= K x { \frac{10^{-8}}{0.0025}\frac{10^{-8}}{0.0225} }

= K x 10-8 x (400 – 44.44)

= 9 x 109 x 10-8 x 355.56

= 32000

= 3.2 x 104 N/C

 

Thus, electric field intensity at point A is 7.2 x 104 N/C directed towards left.

 

Calculating Electric Field Intensity At Point C-

 

Let-

  • Electric field intensity at point C due to charge q1 = E1C
  • Electric field intensity at point C due to charge q2 = E2C

 

 

It is clear that-

  • Electric field at point C due to charge q1 is directed away from q1 along the line joining q1 and C.
  • Electric field at point C due to charge q2 is directed towards q2 along the line joining q2 and C.
  • Angle between \overrightarrow{E_{1C}} and \overrightarrow{E_{2C}} is 120°.

 

Now,

Electric field intensity at point C due to charge q1

= K x ( \frac{q_{1}}{0.1^{2}} )

= K x 10-8 x 100

= 9 x 109 x 10-8 x 100

= 9000 N/C directed away from q1 along the line joining q1 and point C.

 

Similarly,

Electric field intensity at point C due to charge q2

= 9000 N/C directed towards q2 along the line joining q2 and point C.

 

Net electric field at point C

= sqrt { E21C + E22C + 2 x E1C x E2C x cos120° }

= sqrt { (9000)2 + (9000)2 + 2 x 9000 x 9000 x ( \frac{-1}{2} ) }

= sqrt { 2 x 81 x 106 – 81 x 106 }

= sqrt { 81 x 106 }

= 9 x 103 N/C

 

Thus, electric field intensity at point C is 9 x 103 N/C.

The direction of electric field is along the bisector of the angle between \overrightarrow{E_{1C}} and \overrightarrow{E_{2C}} towards right.

This is because \overrightarrow{E_{1C}} and \overrightarrow{E_{2C}} are equal in magnitude.

 

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

Watch this Video Lecture

 

Next Article- Electric Field Lines

 

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