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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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Electric Field | Electric Field Intensity

Electric Field-

 

Electric Field is defined as-

 

The space around an electric charge in which its effect can be experienced is called as its electric field.

OR

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.

 

Types of Electric Field-

 

There are following 4 types of electric field-

 

 

  1. Uniform Electric Field
  2. Non-Uniform Electric Field
  3. Variable Electric Field
  4. Constant Electric Field

 

1. Uniform 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.

 

2. Non-Uniform Electric Field-

 

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.

 

3. Variable Electric Field-

 

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

E = f(t)

 

4. Constant Electric Field-

 

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

E ≠ f(t)

 

Electric Field Intensity-

 

Electric Field Intensity is defined as-

 

Electric field intensity at any point is the strength of electric field at that point.

OR

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

 

If \overrightarrow{F} is the force acting on a small test charge q0 at any point in the electric field, then electric field intensity \overrightarrow{E} at that point is given by-

 

 

It is important to note that-

  • The test charge q0 may have its own electric field.
  • This may modify the electric field of the source charge.
  • So to minimize this effect, we write \overrightarrow{E} as-

 

 

From the above relation,

Force experienced by any point charge q0 in the electric field of intensity \overrightarrow{E} may be calculated as-

\overrightarrow{F} = q0 . \overrightarrow{E}

 

We know F = ma.

So, acceleration of electric charge q0 having mass m in electric field \overrightarrow{E} may be given as-

 

 

Electric Field Intensity Due To Point Charge-

 

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 q0 at point P.

 

 

According to Coulomb’s law,

Force \overrightarrow{F} at point P is given by-

 

 

Also, we know \overrightarrow{E} = \overrightarrow{F} / q0

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.

 

 

Electric Field Intensity Due To System Of Point Charges-

 

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-

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

 

Important Points-

 

Point-01:

 

  • 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.

 

Point-02:

 

  • 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.

 

Point-03:

 

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

 

Point-04:

 

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.

 

Point-05:

 

  • 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.

 

Point-06:

 

A point charge q0 is kept in an electric field of strength E and experiences a force F, then-

  • E < F / q0 if field is diverging
  • E = F / q0 if field is uniform
  • E > F / q0 if field is converging

 

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

Watch this Video Lecture

 

Next Article- Problems On Electric Field

 

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