Question 104 of 180

A capacitor is made of two square plates each of side ‘ $a$ ’ making a very small angle $\alpha$ between them, as shown in figure. The capacitance will be close to:

A$\dfrac{{\in }_{0}{a}^{2}}{d}\left(1-\dfrac{\alpha a}{2d}\right)$B$\dfrac{{\in }_{0}{a}^{2}}{d}\left(1-\dfrac{\alpha a}{4d}\right)$C$\dfrac{{\in }_{0}{a}^{2}}{d}\left(1+\dfrac{\alpha a}{d}\right)$D$\dfrac{{\in }_{0}{a}^{2}}{d}\left(1-\dfrac{3\alpha a}{2d}\right)$

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Answer: (A) $\dfrac{{\in }_{0}{a}^{2}}{d}\left(1-\dfrac{\alpha a}{2d}\right)$

$dc=\dfrac{{\in }_{0}adx}{d+\alpha x}$
$\Rightarrow c=\dfrac{{\in }_{0}a}{\alpha }{\left[In\left(d+\alpha x\right)\right]}_{0}^{a}$
$=\dfrac{{\in }_{0}a}{\alpha }In\left(1+\dfrac{\alpha a}{d}\right)\approx \dfrac{{\in }_{0}{a}^{2}}{d}\left(1-\dfrac{\alpha a}{2b}\right)$

JEE Main Physics practice question on Capacitors (Capacitors with Dielectric), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.