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Two conductors carrying equal and opposite charges create a non-uniform field as shown in fig. What is the capacity of this capacitor if the field along y-axis varies as $E = \frac{Q}{ε_{0} A} [1 + {By}^{2}]$ with B = constant

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$C = \frac{ε_{0} A}{d} 1$

$C = \frac{ε_{0} A}{d [1 + \frac{1}{2} {Bd}^{2}]}$

$C = \frac{ε_{0} AB}{2 d}$

$C = \frac{ε_{0} A}{d [1 + \frac{1}{3} {Bd}^{2}]}$

answer is B.

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Detailed Solution

$\begin{array}{l} - \frac{dV}{dy} = \frac{Q}{ε_{0} A} [1 + {By}^{2}] \\ ⇒ - \int_{V_{A}}^{V_{B}} dV = \frac{Q}{ε_{0} A} \int_{0}^{d} (1 + {By}^{2}) dy \end{array}$

$So, C = \frac{Q}{V} = \frac{ε_{0} A}{d [1 + \frac{1}{3} {Bd}^{2}]}$

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