Answer to Question #138558 in Electric Circuits for Oler

Question

A fully charged parallel-plate capacitor remains connected to a battery while you slide a dielectric between the plates. Do the following quantities increase, decrease or remain the same? (i) C (ii) Q (iii) Change in V (iv) the energy stored in the capacitor

Answer 1

We have $C = \epsilon_0\times A/d$ for free space.

Where $\epsilon_0$ is the permittivity of free space.

$A$ is the area of parallel plate capacitor

$d$ is the distance between parallel plates

When a dielectric with dielectric constant K is inserted the formulae becomes

$C = K \times\epsilon_0\times A/d$

(i) Hence from above equation The capacitance increases by K times.

(ii) We have Charge $Q = C \times V$ Where $Q$ is the charge and $V$ is the voltage across capacitor.

As capacitance increases by K times Charge increases by K times.

(iii) As battery is kept connected There is no change in V i.e change in V =0.

(iv) = Energy = $1/2 \times C \times V^2$ This too is directly proportional to Capacitance. That means

Energy increases by K times