Question

The potentials at points A and B are \( V_A \) and \( V_B \) respectively for the charges \( +q \) and \( -q \) placed at distances ' x ' each from points A and B as shown in figure. The distance between points A and B is ' y '. The net potential (\( V_A - V_B \)) is proportional to

• A. \( \frac{2qy}{x(x+y)} \)
• B. \( \frac{-2qy}{x(x+y)} \)
• C. \( \frac{qy}{x(x+y)} \)
• D. Zero

Hint:

- Potential is scalar → add algebraically - Carefully track distances for each charge

Answer 1

The Correct Option is A

Concept:
Electric potential due to a point charge: \[ V = \frac{kq}{r} \]

Step 1: Write potential at A.
From +q at distance $x$: \[ V_{A1} = \frac{kq}{x} \] From $-q$ at distance $(x+y)$: \[ V_{A2} = \frac{-kq}{x+y} \] \[ V_A = \frac{kq}{x} - \frac{kq}{x+y} \]

Step 2: Write potential at B.
From +q at distance $(x+y)$: \[ V_{B1} = \frac{kq}{x+y} \] From $-q$ at distance $x$: \[ V_{B2} = \frac{-kq}{x} \] \[ V_B = \frac{kq}{x+y} - \frac{kq}{x} \]

Step 3: Find difference.
\[ V_A - V_B = \left(\frac{kq}{x} - \frac{kq}{x+y}\right) - \left(\frac{kq}{x+y} - \frac{kq}{x}\right) \] \[ = \frac{kq}{x} - \frac{kq}{x+y} - \frac{kq}{x+y} + \frac{kq}{x} \] \[ = 2\left(\frac{kq}{x} - \frac{kq}{x+y}\right) \]

Step 4: Simplify.
\[ = 2kq \left( \frac{(x+y) - x}{x(x+y)} \right) = 2kq \left( \frac{y}{x(x+y)} \right) \]

Step 5: Final proportionality.
\[ V_A - V_B \propto \frac{2qy}{x(x+y)} \]