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  •   A capacitor is made of two square plates each of side 'a' making a very small angle

 

A capacitor is made of two square plates each of side 'a' making a very small angle \alpha between them, as shown in the figure. The capacitance will be close to : 
Option: 1 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{4 d } \right )

Option: 2 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 + \frac{\alpha a }{4 d } \right )

Option: 3 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{2 d } \right )

Option: 4 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{3 \alpha a }{2 d } \right )
 

Answers (1)

best_answer

Take an elemental strip of length dx at a distance x from one end

So the capacitance of an elemental strip is \mathrm{dc}=\frac{\varepsilon_{0} \mathrm{ad} \mathrm{x}}{\mathrm{d}+\alpha \mathrm{x}}

So integrating this will give total capacitance=C=\int \mathrm{dc}=\int_{0}^{a}\frac{\varepsilon_{0} \mathrm{ad} \mathrm{x}}{\mathrm{d}+\alpha \mathrm{x}}

\begin{array}{l}{\Rightarrow \quad c=\frac{\varepsilon_{0} a}{\alpha}[\ln (d+\alpha x)]_{0}^{a}} \\ {C=\frac{\varepsilon_{0} a}{\alpha} \ln \left(1+\frac{\alpha a}{d}\right) \approx \frac{\varepsilon_{0} a^{2}}{d}\left(1-\frac{\alpha a}{2 d}\right)}\end{array}  .....(Using \ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots)

 

Hence the correct option is (3).

 

Posted by

vishal kumar

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