A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_{1}$ and dielectric constant $K_{1}$ and the other has thickness $d_{2}$ and dielectric constant $K_{2}$ as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (= $d_{1} + d_{2}$ ) and effective dielectric constant K. The K is

$\frac{K_{1} d_{1} + K_{2} d_{2}}{d_{1} + d_{2}}$
$\frac{K_{1} d_{1} + K_{2} d_{2}}{K_{1} + K_{2}}$
$\frac{K_{1} K_{2} (d_{1} + d_{2})}{{K_{2} d}_{1} + K_{1} d_{2}}$
$\frac{{2 K}_{1} K_{2}}{K_{1} + K_{2}}$

Question

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness $d_{1}$ and dielectric constant $K_{1}$ and the other has thickness $d_{2}$ and dielectric constant $K_{2}$ as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (= $d_{1} + d_{2}$ ) and effective dielectric constant K. The K is

$\frac{K_{1} d_{1} + K_{2} d_{2}}{d_{1} + d_{2}}$
$\frac{K_{1} d_{1} + K_{2} d_{2}}{K_{1} + K_{2}}$
$\frac{K_{1} K_{2} (d_{1} + d_{2})}{{K_{2} d}_{1} + K_{1} d_{2}}$
$\frac{{2 K}_{1} K_{2}}{K_{1} + K_{2}}$

Toppr Admin · Accepted Answer

The capacities of two individual condensers areC1-K1-x3B5-0Ad1-C2-K2-x3B5-0Ad2The arrangement is equivalent to two capacitors joined in series-x2234- Equivalent capacitance- 1Ceq-1C1-1C2-d1K1-x3B5-0A-d2K2-x3B5-0A-1-x3B5-0A-d1K1-d2K2-1-x3B5-0A-K2d1-K1d2K1-K2-or Ceq-x3B5-0A-K1-K2K2d1-K1d2-i-Also Ceq-K-x3B5-0Ad1-d2From -i- and -ii- -x3B5-0A-K1-K2K2d1-K1d2-x3B5-0A-Kd1-d2-x2234-K-K1K2-d1-d2-K1d2-K2d1