A) \[\frac{1}{2}\,\frac{k\,{{\in }_{0}}{{a}^{2}}}{d}\]
B) \[\frac{k\,{{\in }_{0}}{{a}^{2}}}{d}\,\text{In}\,K\]
C) \[\frac{k\,{{\in }_{0}}{{a}^{2}}}{d\,(K-1)}\,\text{In}\,k\]
D) \[\frac{k\,{{\in }_{0}}{{a}^{2}}}{2d(K+1)}\]
Correct Answer: C
Solution :
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\[\frac{y}{x}\,=\,\frac{d}{a}\] |
\[dy=\,\frac{\text{d}}{a}\,(\text{d}x)\] |
\[\frac{1}{dc}\,=\,\frac{y}{\text{KE}.adx}+\frac{(d-y)}{{{\in }_{0}}\,adx}\] |
\[\frac{1}{dc}\,=\,\frac{1}{{{\in }_{0}}}adx\left( \frac{y}{k}+d-y \right)\] |
\[\int{dc}\,=\,\int{\frac{{{\varepsilon }_{0}}adx}{\frac{y}{k}+d-y}}\] |
\[c={{\varepsilon }_{0}}a.\frac{a}{d}\int_{0}^{d}{\frac{dy}{d+y\left( \frac{1}{k}-1 \right)}}\] |
\[=\frac{{{\varepsilon }_{0}}{{a}^{2}}}{\left( \frac{1}{k}\,-\,1 \right)}\left[ \text{In}\left( d+y\left( \frac{1}{k}\,-\,1 \right) \right) \right]_{0}^{\text{d}}\] |
\[=\,\frac{k{{\varepsilon }_{0}}{{a}^{2}}}{(1\,-\,k)d}\,\text{In}\left( \frac{d+d\left( \frac{1}{k}\,-\,1 \right)}{d} \right)\] |
\[=\,\frac{k{{\varepsilon }_{0}}{{a}^{2}}}{(1-k)d}\,\text{In}\,\left( \frac{1}{k} \right)\,=\,\frac{k{{\varepsilon }_{0}}{{a}^{2}}\,\text{In}\,k}{(k\,-\,1)d}.\] |
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