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 :

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) = $$x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - ...$$

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: