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Question: Separation between the plates of a parallel plate capacitor is ![](https://cdn.pureessence.tech/canv...

Separation between the plates of a parallel plate capacitor is and the area of each plate is AA . When a slab of material of dielectric constant kk and thickness t(t<d)t ( t < d ) is introduced between the plates, its capacitance becomes

A

ε0Ad+t(1−1k)\frac { \varepsilon _ { 0 } A } { d + t \left( 1 - \frac { 1 } { k } \right) }

B

ε0Ad+t(1+1k)\frac { \varepsilon _ { 0 } A } { d + t \left( 1 + \frac { 1 } { k } \right) }

C

ε0Ad−t(1−1k)\frac { \varepsilon _ { 0 } A } { d - t \left( 1 - \frac { 1 } { k } \right) }

D

ε0Ad−t(1+1k)\frac { \varepsilon _ { 0 } A } { d - t \left( 1 + \frac { 1 } { k } \right) }

Answer

ε0Ad−t(1−1k)\frac { \varepsilon _ { 0 } A } { d - t \left( 1 - \frac { 1 } { k } \right) }

Explanation

Solution

Potential difference between the plates V

= Vair + Vmedium

=σε0×(d−t)+σKε0×t= \frac { \sigma } { \varepsilon _ { 0 } } \times ( d - t ) + \frac { \sigma } { K \varepsilon _ { 0 } } \times t

⇒ V=σε0(d−t+tK)V = \frac { \sigma } { \varepsilon _ { 0 } } \left( d - t + \frac { t } { K } \right)

=QAε0(d−t+tK)= \frac { Q } { A \varepsilon _ { 0 } } \left( d - t + \frac { t } { K } \right)

Hence capacitance C=QV=QQAε0(d−t+tK)C = \frac { Q } { V } = \frac { Q } { \frac { Q } { A \varepsilon _ { 0 } } \left( d - t + \frac { t } { K } \right) }

=ε0A(d−t+tK)=ε0Ad−t(1−1K)= \frac { \varepsilon _ { 0 } A } { \left( d - t + \frac { t } { K } \right) } = \frac { \varepsilon _ { 0 } A } { d - t \left( 1 - \frac { 1 } { K } \right) }