A parallel plate capacitor having cross-secti | vedclass

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(C) The original capacitance of the parallel plate capacitor with air is given by $C_{0} = \frac{\varepsilon_{0} A}{d}$.When a dielectric slab of thic

Vedclass · Accepted Answer

$8:5$

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(C) The original capacitance of the parallel plate capacitor with air is given by $C_{0} = \frac{\varepsilon_{0} A}{d}$.When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,the new capacitance $C$ is given by the formula:$C = \frac{\varepsilon_{0} A}{d - t + \frac{t}{K}}$Given $t = \frac{d}{2}$ and $K = 4$,we substitute these values into the formula:$C = \frac{\varepsilon_{0} A}{d - \frac{d}{2} + \frac{d/2}{4}}$$C = \frac{\varepsilon_{0} A}{\frac{d}{2} + \frac{d}{8}}$$C = \frac{\varepsilon_{0} A}{\frac{4d + d}{8}} = \frac{\varepsilon_{0} A}{\frac{5d}{8}}$$C = \frac{8}{5} \frac{\varepsilon_{0} A}{d} = \frac{8}{5} C_{0}$Therefore,the ratio of the new capacitance to the original capacitance is $\frac{C}{C_{0}} = \frac{8}{5}$.

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