Question

A uniform electric field vector \(\vec{E}\) exists along horizontal direction as shown. The electric potential at A is V_A . 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

• A. q[V_A + Ex]
• B. q[Ex - V_A]
• C. qEx
• D. q[V_A - Ex]

Answer 1

The Correct Option is D

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.

Answer 2

The electric potential energy \( U \) of a point charge \( q \) in an electric field \( \mathbf{E} \) is given by: \[ U = qV \] where \( V \) is the electric potential at the position of the charge. In this case, a uniform electric field \( \mathbf{E} \) exists along the horizontal direction, and a point charge \( q \) is moved from point \( A \) to point \( B \). The potential at point \( A \) is given as \( V_A \), and the electric field is uniform, meaning the potential varies linearly with distance. The work done in moving the charge in the electric field is related to the change in potential energy, which can be written as: \[ \Delta U = q(V_B - V_A) \] Since the electric field is uniform, the potential difference \( V_B - V_A \) is related to the distance between points \( A \) and \( B \) by: \[ V_B - V_A = -E \cdot d \] where \( d \) is the distance between \( A \) and \( B \) and \( E \) is the magnitude of the electric field. Therefore, the potential energy of the charge at point \( B \) is: \[ U_B = q(V_A - E \cdot d) \]

Thus, the potential energy at point \( B \) is: \[{q[V_A - E \cdot x]} \]

\(\textbf{Correct Answer:}\) (D) \( q[V_A - Ex] \)