JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 7)
A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :
$$\varepsilon (x) = {\varepsilon _0} + kx$$, for $$\left( {0 < x \le {d \over 2}} \right)$$
$$\varepsilon (x) = {\varepsilon _0} + k(d - x)$$, for $$\left( {{d \over 2} \le x \le d} \right)$$
$$\varepsilon (x) = {\varepsilon _0} + kx$$, for $$\left( {0 < x \le {d \over 2}} \right)$$
$$\varepsilon (x) = {\varepsilon _0} + k(d - x)$$, for $$\left( {{d \over 2} \le x \le d} \right)$$
$${\left( {{\varepsilon _0} + {{kd} \over 2}} \right)^{2/kA}}$$
$${{kA} \over {2\ln \left( {{{2{\varepsilon _0} + kd} \over {2{\varepsilon _0}}}} \right)}}$$
0
$${{kA} \over 2}\ln \left( {{{2{\varepsilon _0}} \over {2{\varepsilon _0} - kd}}} \right)$$
Explanation
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Taking an element of width dx at a distance x(x < d/2) from left plate
$$dc = {{({\varepsilon _0} + kx)A} \over {dx}}$$
Capacitance of half of the capacitor
$${1 \over C} = \int\limits_0^{d/2} {{1 \over {dc}} = {1 \over A}\int\limits_0^{d/2} {{{dx} \over {{\varepsilon _0} + kx}}} } $$
$${1 \over C} = {1 \over {kA}}\ln \left( {{{{\varepsilon _0} + kd/2} \over {{\varepsilon _0}}}} \right)$$
Capacitance of second half will be same
$${C_{eq}} = {C \over 2} = {{kA} \over {2\ln \left( {{{2{\varepsilon _0} + kd} \over {2{\varepsilon _0}}}} \right)}}$$
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