**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 q_{0} having mass m in an electric field having electric field intensity 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-

** = + + + ……. + **

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 10^{3} N/C

Thus, electric field intensity is 1.5 x 10^{3} 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/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.

**Problem-03:**

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.

**Solution-**

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

**Step-01: Calculating Acceleration Of Electron-**

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.

**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 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.

**Problem-04:**

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.

**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 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

**Step-01: Calculating Acceleration Of Electron-**

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}

**Step-02: Calculating Time Of Travel-**

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.

**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 (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}.

**Problem-06:**

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-

**Solution-**

**Calculating Electric Field Intensity At Point A-**

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.

**Calculating Electric Field Intensity At Point B-**

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.

**Calculating Electric Field Intensity At Point C-**

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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