When a positive `q` charge is taken from lower potential to a higher potential point, then its potential energy will

Options

A. remin the same
B. increase
C. decrease
D. become zero

Answer

To solve the problem of how the potential energy of a positive charge \( q \) changes when it is moved from a lower potential to a higher potential, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Potentials**:
- Let the initial potential at point \( Vi \) be lower than the final potential at point \( Vf \). Thus, we have \( Vf > Vi \).

2. **Define the Potential Energy**:
- The potential energy \( U \) of a charge \( q \) at a potential \( V \) is given by the formula:
\[ U = qV \]
- Therefore, the potential energy at the initial point \( Vi \) is:
\[ U_i = q \cdot Vi \]
- And the potential energy at the final point \( Vf \) is:
\[ U_f = q \cdot Vf \]

3. **Calculate the Change in Potential Energy**:
- The change in potential energy \( \Delta U \) when moving from \( Vi \) to \( Vf \) is given by:
\[ \Delta U = U_f - U_i = q \cdot Vf - q \cdot Vi \]
- This simplifies to:
\[ \Delta U = q(Vf - Vi) \]

4. **Analyze the Sign of the Change**:
- Since it is given that \( Vf > Vi \), the term \( (Vf - Vi) \) is positive.
- Therefore, \( \Delta U \) becomes:
\[ \Delta U = q \cdot (Vf - Vi) > 0 \]
- Since \( q \) is a positive charge, the change in potential energy \( \Delta U \) is positive.

5. **Conclusion**:
- Since the change in potential energy \( \Delta U \) is positive, we conclude that the potential energy of the positive charge \( q \) increases when it is moved from a lower potential to a higher potential.

### Final Answer:
The potential energy will **increase**.

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