Question

Two parallel plate capacitors of capacity $C$ and $3 C$ are connected in parallel combination and charged to a potential difference $18\, V$ The battery is then disconnected and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $9$ The final potential difference across the combination of capacitors will be______ $V$

Answer 1

Correct Answer: 6

Initially, the capacitors of capacitance \(C_1=C\) and \(C_2=3C\) are connected in parallel to a potential difference \(V=18\,V\). The charge on each capacitor is given by \(Q_1=C_1V=CV\) and \(Q_2=C_2V=3CV\). Therefore, the total initial charge is \(Q_{\text{initial}}=Q_1+Q_2=CV+3CV=4CV\).

When the dielectric material with a constant \(k=9\) is introduced into capacitor \(C_1\), its capacitance becomes \(C_1' = kC = 9C\).

The new equivalent capacitance of the capacitors in parallel is \(C_{\text{new}}=C_1'+C_2=9C+3C=12C\).

Since the battery is disconnected, the total charge remains conserved: \(Q_{\text{final}}=Q_{\text{initial}}\). Thus, the final voltage \(V_f\) across the capacitors is given by \(V_f=\frac{Q_{\text{final}}}{C_{\text{new}}}=\frac{4CV}{12C}=\frac{4 \times 18\,\text{V}}{12}=6\,\text{V}\).

Thus, the final potential difference across the combination of capacitors is \(6\,V\), which falls within the given range of 6,6.