Question

Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.

Hint:

Always check battery connection status:
Battery CONNECTED $\implies V$ remains constant $\implies Q' = KQ$, $U' = KU$.
Battery DISCONNECTED $\implies Q$ remains constant $\implies V' = V/K$, $U' = U/K$.

Answer 1

Concept:
• When capacitors remain connected to a DC battery of potential $V$, potential difference across plates remains constant ($V' = V$).

• Insertion of dielectric slab of dielectric constant $K$ increases capacitance from $C$ to $C' = K C$.

Step 1: Effect on Capacitance
Initial capacitances are $C_1$ and $C_2$.
When dielectric slabs are inserted, new capacitances become:
\[ C_1' = K C_1 \quad \text{and} \quad C_2' = K C_2 \]
Thus, capacitance of each capacitor increases by factor $K$.

Step 2: (i) Effect on Charge on Each Capacitor
Since battery remains connected, potential difference $V$ across each capacitor remains unchanged.
Initial charges: $Q_1 = C_1 V$ and $Q_2 = C_2 V$.
New charges after dielectric insertion:
\[ Q_1' = C_1' V = (K C_1) V = K Q_1 \]
\[ Q_2' = C_2' V = (K C_2) V = K Q_2 \]
Therefore, charge on each capacitor increases $K$ times.

Step 3: (ii) Effect on Stored Energy
Initial energy stored in capacitors:
\[ U_1 = \frac{1}{2} C_1 V^2 \quad \text{and} \quad U_2 = \frac{1}{2} C_2 V^2 \]
New energy stored in capacitors:
\[ U_1' = \frac{1}{2} C_1' V^2 = \frac{1}{2} (K C_1) V^2 = K U_1 \]
\[ U_2' = \frac{1}{2} C_2' V^2 = \frac{1}{2} (K C_2) V^2 = K U_2 \]
Therefore, energy stored in each capacitor increases $K$ times.

Step 4: Conclusion
When battery remains connected:
(i) Charge on each capacitor increases to $K$ times its initial value.
(ii) Stored energy in each capacitor increases to $K$ times its initial value.