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 :

Question

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

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

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

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

Solution

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