A uniform electric field vector \(\vec{E}\) exists along horizontal direction as shown. The electric potential at A is VA . A small point charge q is slowly taken from A to B along the curved path as shown. The potential energy of the charge when it is at point B is
- q[VA + Ex]
- q[Ex - VA]
- qEx
- q[VA - Ex]
The Correct Option is D
Solution and Explanation
Given:
- A uniform electric field \( \vec{E} \) exists along the horizontal direction.
- The electric potential at point A is \( V_A \).
- A small charge \( q \) is moved slowly from point A to point B along a curved path.
Step 1: Expression for Potential Energy
The potential energy of the charge at point B is given by:
\[ U_B = q (V_A - Ex) \]
Answer: The correct option is D.
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