Question

Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is $ V $. The potential difference between the points A and B (shown in the figure) is:

• A. \( \frac{1}{4} V \)
• B. \( \frac{2}{5} V \)
• C. \( \frac{3}{4} V \)
• D. \( 1 V \)

Hint:

To calculate potential difference in a parallel plate setup, use the relationship \( V_{AB} = \frac{E \times d}{V} \), where \( d \) is the distance between the points of interest.

Answer 1

The Correct Option is B

Given a potential difference \( V \) between two plates separated by 10 cm. Points A and B are located at distances of 3 cm and 4 cm, respectively, from one of the plates, such that the total separation between the plates is 10 cm. The relationship between potential difference, electric field (\( E \)), and distance (\( \Delta d \)) is \( \Delta V = E \Delta d \). Therefore, the electric field between the plates is \( E = \frac{V}{10 \, \text{cm}} \).

The potential difference between points A and B is calculated as: \[ V_{AB} = E \times 4 \, \text{cm} \] Substituting the value of \( E \): \[ V_{AB} = \frac{V}{10 \, \text{cm}} \times 4 \, \text{cm} = \frac{2V}{5} \] Consequently, the potential difference between points A and B is \( \frac{2}{5} V \).