JEE MAIN - Physics (2025 - 2nd April Evening Shift - No. 12)
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 :_2nd_April_Evening_Shift_en_12_1.png)
1 V
$\frac{1}{4} \mathrm{~V}$
$\frac{3}{4} \mathrm{~V}$
$\frac{2}{5} \mathrm{~V}$
Explanation
_2nd_April_Evening_Shift_en_12_2.png)
$$\begin{aligned} & \text { Using } \Delta V=E(\Delta d) \\ & V=E(10) \\ & V_{A B}=E \cdot 4=\frac{V}{10} \times 4=\frac{2 V}{5} \end{aligned}$$
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