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:
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:
Show Hint
- \( \frac{1}{4} V \)
- \( \frac{2}{5} V \)
- \( \frac{3}{4} V \)
- \( 1 V \)
The Correct Option is B
Solution and Explanation
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 \( \frac{2}{5} V \).
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