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A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

1 Answer

Let $E_0 = V_0/d$ be the electric field between the plates when there is no dielectric and the potential difference is V0. If the dielectric is now inserted, the electric field in the dielectric will be $E = E_0/K. $
The potential difference will then be
$V= E_0 \bigg( \large\frac{1}{4} d \bigg)+ \large\frac{e_0}{K} \bigg( \large\frac{3}{4} $$d\bigg)$
$\quad= E_0 d \bigg( \large\frac{1}{4} +\frac{3}{4K} \bigg) =v_0 \large\frac{E+3}{4K}$
The potential difference decreases by the factor $(K + 3)/4K$ while the free charge $Q_0$ on the plates remains unchanged. The capacitance thus increases
$C= \large\frac{Q_0}{V} =\frac{4K}{K+3} \frac{Q_0}{V_0}=\frac{4 K}{K+3}$$C_0$
answered Jun 13, 2014 by meena.p
 

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