Question 168: Electrostats & Capacitance

Question

A slab of dielectric constant $$\mathrm{K}$$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $$\frac{3}{4} \mathrm{~d}$$, where $$\mathrm{d}$$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :
(Given $$\mathrm{C}_{0}$$ = capacitance of capacitor with air as medium between plates.)

Accepted Answer

$${C_0} = {{{\varepsilon _0}A} \over d}$$
$$C = {{{\varepsilon _0}A} \over {d - {{3d} \over 4} + {{3d} \over {4K}}}} = {{4{\varepsilon _0}AK} \over {3d + Kd}}$$
$$ = {{4K{C_0}} \over {3 + K}}$$

Answer

$$\frac{4 K C_{0}}{3+K}$$

Answer

$$\frac{3 K C_{0}}{3+K}$$

Answer

$$\frac{3+K}{4 K C_{0}}$$

Answer

$$\frac{K}{4+K}$$

Electrostats & Capacitance

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