Question 315: Electrostatics & Capacitance

Question

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 :
![](https://imagex.cdn.examgoal.net/7k9k2k5hh3y6k/633a7668-68e0-4ee9-949f-53e48a54fae9/2547a0c0-38c7-11ea-92dd-4b9c3519996f/file-7k9k2k5hh3y6l-341w.jpg)

Accepted Answer

dC = ${{{\varepsilon _0}adx} \over {d + \alpha x}}$
$ 	herefore $ C = $\int\limits_0^a {{{{\varepsilon _0}adx} \over {d + \alpha x}}} $
= ${{{\varepsilon _0}a} \over \alpha }\left[ {\ln \left( {d + \alpha } ight)} ight]_0^a$
= ${{{\varepsilon _0}a} \over \alpha }\ln \left[ {{{d + \alpha a} \over d}} ight]$
= ${{{\varepsilon _0}a} \over \alpha }\ln \left[ {1 + {{\alpha a} \over d}} ight]$
= ${{{\varepsilon _0}a} \over \alpha }\left( {{{\alpha a} \over d} - {{{\alpha ^2}{a^2}} \over {{d^2}}}} ight)$
= ${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} ight)$
Note :
ln(1 + x) = $x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - ...$
![](https://imagex.cdn.examgoal.net/7kd4dk6a323cm/b98f902c-de3f-4c07-89e2-308649601e23/ecadc160-4882-11ea-b62d-51ccc5d4afb6/file-7kd4dk6a323cn-406w.jpg)

Answer

${{{\varepsilon _0}{a^2}} \over d}\left( {1 + {{\alpha a} \over {d}}} \right)$

Answer

${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {4d}}} \right)$

Answer

${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)$

Answer

${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{3\alpha a} \over {2d}}} \right)$

Electrostatics & Capacitance

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 :

Choose the correct answer: