JEE MAIN - Physics (2024 - 9th April Morning Shift - No. 12)
A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure. If the area of each stair is $$\frac{A}{3}$$ and the height is $$d$$, the capacitance of the arrangement is :
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$$\frac{11 \epsilon_{\mathrm{o}} \mathrm{A}}{20 \mathrm{~d}}$$
$$\frac{13 \epsilon_{\mathrm{o}} \mathrm{A}}{17 \mathrm{~d}}$$
$$\frac{18 \epsilon_{\mathrm{o}} \mathrm{A}}{11 \mathrm{~d}}$$
$$\frac{11 \epsilon_{\mathrm{o}} \mathrm{A}}{18 \mathrm{~d}}$$
Explanation
$$\begin{aligned}
& C=C_1+C_2+C_3 \\
& =\frac{A \epsilon_0}{3 d}+\frac{A}{6} \frac{\epsilon_0}{d}+\frac{A \epsilon_0}{g D} \\
& =\frac{A \epsilon_0}{d}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right) \\
& =\frac{11}{18} \frac{A \epsilon_0}{d}
\end{aligned}$$
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