The electric potential at point A is 1V and at another point B is 5V. A charge `3muC` is released from B. What will be the kinetic energy of the charge as it passes through A?

Options

A. `8xx10^(-6)J`
B. `12xx10^(-6)J`
C. `12xx10^(-9)J`
D. `4xx10^(-6)J`

Explanation

To solve the problem, we need to find the kinetic energy of a charge as it moves from point B to point A, given the electric potentials at those points.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Electric potential at point A, \( V_A = 1 \, \text{V} \)
- Electric potential at point B, \( V_B = 5 \, \text{V} \)
- Charge, \( Q = 3 \, \mu\text{C} = 3 \times 10^{-6} \, \text{C} \)

2. **Calculate the Change in Electric Potential:**
- The change in electric potential, \( \Delta V \), as the charge moves from B to A is given by:
\[ \Delta V = V_A - V_B = 1 \, \text{V} - 5 \, \text{V} = -4 \, \text{V} \]

3. **Use the Work-Energy Principle:**
- The work done on the charge by the electric field is equal to the change in kinetic energy:
\[ \Delta KE = Q \cdot \Delta V \]

4. **Substitute the Values:**
- Now substituting the values into the equation:
\[ \Delta KE = (3 \times 10^{-6} \, \text{C}) \cdot (-4 \, \text{V}) = -12 \times 10^{-6} \, \text{J} \]

5. **Interpret the Result:**
- The negative sign indicates that the kinetic energy has increased as the charge moves from a higher potential (B) to a lower potential (A). Therefore, the kinetic energy gained by the charge is:
\[ KE = 12 \times 10^{-6} \, \text{J} = 12 \, \mu\text{J} \]

### Final Answer:
The kinetic energy of the charge as it passes through point A is \( 12 \, \mu\text{J} \).

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