In moving a unit positively charged body from point A to point B external work donw is 40 J and body acquires a kinetic energy of 20 J . The potential differnece `V_(B)- V_(A) `, is

Options

A. 20V
B. `-20V`
C. 40V
D. `-40V`

Answer

To solve the problem, we need to find the potential difference \( V_B - V_A \) when a unit positively charged body is moved from point A to point B. We are given the external work done and the kinetic energy acquired by the body.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Work done by external agent, \( W = 40 \, \text{J} \)
- Change in kinetic energy, \( KE = 20 \, \text{J} \)
- Charge of the body, \( Q = 1 \, \text{C} \) (since it's a unit positively charged body)

2. **Use the Work-Energy Principle:**
The work done by the external agent is equal to the change in mechanical energy, which includes both the change in potential energy and the change in kinetic energy:
\[ W = \Delta PE + \Delta KE \]

3. **Calculate Change in Potential Energy:**
Rearranging the equation gives:
\[ \Delta PE = W - \Delta KE \]
Substituting the known values:
\[ \Delta PE = 40 \, \text{J} - 20 \, \text{J} = 20 \, \text{J} \]

4. **Relate Change in Potential Energy to Potential Difference:**
The change in potential energy (\( \Delta PE \)) can also be expressed in terms of potential difference:
\[ \Delta PE = Q \cdot (V_B - V_A) \]
Substituting \( Q = 1 \, \text{C} \):
\[ 20 \, \text{J} = 1 \, \text{C} \cdot (V_B - V_A) \]

5. **Solve for the Potential Difference:**
This simplifies to:
\[ V_B - V_A = 20 \, \text{V} \]

### Final Answer:
The potential difference \( V_B - V_A \) is \( 20 \, \text{V} \).