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

We are given the potential difference between two plates as \( V \), and the separation between the plates is 10 cm. The distance between points A and B is 3 cm and 4 cm, respectively, with the total distance between the plates being 10 cm. Using \( \Delta V = E \Delta d \), where \( E \) is the electric field and \( \Delta d \) is the distance: \[ V = E \times 10 \, \text{cm} \] From the diagram, we know that \( E = \frac{V}{10} \). 
The potential difference between points A and B is: \[ V_{AB} = E \times 4 \, \text{cm} = \frac{V}{10} \times 4 = \frac{2V}{5} \] Thus, the potential difference between points A and B is \( \frac{2}{5} V \).