A parallel-plate capacitor with plate area $A$ has separation $d$ between the plates. Two dielectric slabs of dielectric constant $K_{1}$ and $K_{2}$ of same area $\frac{A}{2}$ and thickness $\frac{d}{2}$ are inserted in the space between the plates. The capacitance of the capacitor will be given by :

Question

Aicharya · Accepted Answer

C eq = C 1 + C 2 C 3 C 2 + C 3 C eq = A &#949; 0 2 &#160; d + K 1 &#160; A &#949; 0 &#160; d &#215; K 2 &#160; A &#949; 0 &#160; d K 1 &#160; A &#949; 0 &#160; d + K 2 &#160; A &#949; 0 &#160; d C eq = A &#949; 0 2 &#160; d + A &#949; 0 &#160; d K 1 &#160; K 2 &#160; K 1 + K 2 = A &#949; 0 &#160; d 1 2 + K 1 &#160; K 2 &#160; K 1 + K 2

A parallel-plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of…

Solution