Question
A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws $25\%$ more charge from the battery. The dielectric constant of mica is $\_\_\_\_$ .
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Solution
$${C = \frac{\epsilon_{0}\text{ }A}{\text{ }d} }{Q_{1} = CV }$$
$${Q_{2} = \left( c_{eq} \right)v }{Q_{2} = 1.25cv }$$
$$c_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} = \frac{\frac{\epsilon_{0}A}{3} \times \frac{K\epsilon_{0}A}{2}}{\frac{\epsilon_{0}A}{3} + \frac{K\epsilon_{0}A}{2}}$$
$${c_{\text{eq~}} = \frac{\left( \epsilon_{0}A \right)^{2}\left( \frac{K}{6} \right)}{\epsilon_{0}A\left( \frac{2 + 3K}{6} \right)} \Rightarrow C_{\text{eq~}} = \frac{K\epsilon_{0}A}{2 + 3K} }{1.25 \times \frac{\in_{0}\text{ }A}{5} = \frac{K \in_{0}\text{ }A}{2 + 3\text{ }K} \Rightarrow 0.25(2 + 3\text{ }K) = K}$$
$$2 + 3\text{ }K = 4\text{ }K \Rightarrow \text{ }K = 2$$
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