JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 1)
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 :
_8th_January_Evening_Slot_en_1_1.png)
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 + {{\alpha a} \over {d}}} \right)$$
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {4d}}} \right)$$
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)$$
$${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{3\alpha a} \over {2d}}} \right)$$
Explanation
_8th_January_Evening_Slot_en_1_2.png)
dC = $${{{\varepsilon _0}adx} \over {d + \alpha x}}$$
$$ \therefore $$ C = $$\int\limits_0^a {{{{\varepsilon _0}adx} \over {d + \alpha x}}} $$
= $${{{\varepsilon _0}a} \over \alpha }\left[ {\ln \left( {d + \alpha } \right)} \right]_0^a$$
= $${{{\varepsilon _0}a} \over \alpha }\ln \left[ {{{d + \alpha a} \over d}} \right]$$
= $${{{\varepsilon _0}a} \over \alpha }\ln \left[ {1 + {{\alpha a} \over d}} \right]$$
= $${{{\varepsilon _0}a} \over \alpha }\left( {{{\alpha a} \over d} - {{{\alpha ^2}{a^2}} \over {{d^2}}}} \right)$$
= $${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)$$
Note : ln(1 + x) = $$x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - ...$$
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