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InterviewSolution
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A mica slab of thickness equal to the distance between the two plates of a parallel plate air capacitor is inserted in the space between the plates. Explain the changes in capacitance in the following case, When the mica slab is inserted partially. |
Answer» Solution :Fig.4.47 shows the partial INSERTION of the mica slab in the SPACE between the plates of the PARALLEL plate air capacitor. Let the area of the plates of the capacitor be `alpha`, the area of the mica slab inside the capacitor be `alpha_1(alpha_1 lt alpha)`, distance between the plates of the capacitor be d and the DIELECTRIC constant of mica be k.  Therefore, TOTAL capacitance of the capacitor, C = capacitance of parallel plate capacitor of area `alpha_1` with mica as dielectric + capacitance of parallel plate air capacitor of area `(alpha- alpha_1)` `= (in_0 k alpha_1)/(d)+(in_0(alpha-alpha_1))/(d)= (in_0 alpha)/(d)+(in_0 alpha_1(k-1))/(d)` So due to partial insertion of the mica slab, the capacitance increases by `(in_0alpha_1(k-1))/(d)`. |
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