Tag: Electric Field Lines Problems

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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