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:
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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.
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 \( \frac{2}{5} V \).
