A capacitor when filled with a dielectric \(\mathrm{K}=3\) has charge \(\mathrm{Q}_{0}\), voltage…
A capacitor when filled with a dielectric \(\mathrm{K}=3\) has charge \(\mathrm{Q}_{0}\), voltage \(\mathrm{V}_{0}\) and field \(\mathrm{E}_{0}\). If the dielectric is replaced with another one having \(\mathrm{K}=9\) after disconnecting it with the battery the new values of charge, voltage and field will be respectively
When there is no battery, charge remains same while potential difference and electric field decreases
i.e. \(Q^{\prime}=Q_{0}, V^{\prime}=\frac{V_{0} \times 3}{9}=\frac{V_{0}}{3}\) and
\(E^{\prime}=\frac{E_{0} \times 3}{9}=\frac{E_{0}}{3}\)
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