Question

A capacitor when filled with a dielectric \(k=3\) has charge \(Q_0\), voltage \(V_0\) and field \(E_0\). If the dielectric is replaced with another one having \(k=9\), the new values of charge, voltage and field will be respectively

Answer 1

Charge of capacitor remains the same as the capacitor is isolated ( charge is conserved )

After inserting dielectric of dielectric constant \(k\), capacitance becomes \(kC\) where \(C\) is capacitance without dielectric .

After dielectric insertion,
Voltage of capacitor \(V=\dfrac{Q}{C}\) will become \(\dfrac{V}{k}\)

Also , electric field between plates
\(E=\dfrac{V}{d}, (d=\) distance between plates), will become \(\dfrac{E}{k}\)

let \(V\) is voltage, \(Q\) be the charge and \(C\) is capacitance in absence of dielectric. (\(Q=CV\))
with dielectric \(K=3\) charge \(Q=Q_0\),
Voltage \(V_0=\dfrac{V}{3}\) and field \(E_0=\dfrac{E}{3}\)

Let \({V}'\) and \({E}'\) be the new voltage and electric field after inserting dielectric of dielectric constant \(K=9\)
\({V}'=\dfrac{V}{9}=\dfrac{V_0\times~3}{9}=\dfrac{V_0}{3}\)
\({E}'=\dfrac{E}{9}=\dfrac{E_0\times~3}{9}=\dfrac{E_0}{3}\)
Hence, correct option is (D).