A point charge `20 muC` is shifted from infinity to a point P in an electric field with zero acceleration. If the potential of that point is 1000 volt, then
(i) Find out work done by electric field ?
(iii) If the kinetic energy of charge particle is found to increase by 10 mJ when it is brought from infinity to point P, then what is the total work done by external agent ?
(iv) What is the work done by electric field in the part (iii)
(v) If a point charge `30 muC` is released at rest at point P, then find out its kinetic energy at a large distance ?

Answer

Let's solve the problem step by step.

### Given:
- Charge, \( Q = 20 \, \mu C = 20 \times 10^{-6} \, C \)
- Electric potential at point P, \( V = 1000 \, V \)
- Change in kinetic energy when brought from infinity to point P, \( \Delta KE = 10 \, mJ = 10 \times 10^{-3} \, J \)

### (i) Work done by the electric field

The work done \( W \) by the electric field when moving a charge from infinity to a point in an electric field is given by the formula:

\[ W = Q \cdot V \]

Substituting the values:

\[ W = (20 \times 10^{-6} \, C) \cdot (1000 \, V) = 20 \times 10^{-3} \, J = 0.02 \, J \]

### (ii) Total work done by the external agent

The total work done by the external agent can be calculated using the change in kinetic energy and the work done by the electric field. The total work done by the external agent \( W_{ext} \) is given by:

\[ W_{ext} = \Delta KE + W \]

Substituting the values:

\[ W_{ext} = (10 \times 10^{-3} \, J) + (20 \times 10^{-3} \, J) = 30 \times 10^{-3} \, J = 0.03 \, J \]

### (iii) Work done by the electric field in part (ii)

The work done by the electric field is equal to the negative of the total work done by the external agent, since they act in opposite directions:

\[ W_{electric} = -W_{ext} = -30 \, mJ = -0.03 \, J \]

### (iv) Kinetic energy of the charge at a large distance

When a charge is released from rest at point P, its potential energy is converted into kinetic energy as it moves to a large distance. The initial potential energy \( PE \) at point P is given by:

\[ PE = Q \cdot V \]

For the charge \( 30 \, \mu C \):

\[ PE = (30 \times 10^{-6} \, C) \cdot (1000 \, V) = 30 \times 10^{-3} \, J = 0.03 \, J \]

At a large distance, all this potential energy will convert into kinetic energy \( KE \):

\[ KE = PE = 0.03 \, J \]

### Summary of Answers:
1. Work done by electric field: \( 0.02 \, J \)
2. Total work done by external agent: \( 0.03 \, J \)
3. Work done by electric field in part (ii): \( -0.03 \, J \)
4. Kinetic energy at a large distance: \( 0.03 \, J \)