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 capacitors be affected after the slab is introduced?

Hint:

For a capacitor connected to a battery: \[ V=\text{constant} \] When a dielectric is inserted, \[ C' = KC \] Therefore, \[ Q'=KQ \] and \[ U'=KU \] Remember that charge and energy increase because the battery supplies additional charge to maintain the same voltage.

Answer 1

Concept: When a dielectric slab is inserted completely between the plates of a capacitor, the capacitance increases by a factor equal to the dielectric constant \(K\). If the capacitor remains connected to a battery, the potential difference across the capacitor remains constant because the battery continuously maintains the same voltage. Therefore, \[ C' = KC \] where
• \(C\) = original capacitance,
• \(C'\) = new capacitance after inserting dielectric,
• \(K\) = dielectric constant of the material. Since the capacitors remain connected to the battery, the voltage across each capacitor remains unchanged. \[ V=\text{constant} \]

Step 1: Initial charge on the capacitors Before inserting the dielectric slab: For capacitor \(C_1\), \[ Q_1=C_1V \] For capacitor \(C_2\), \[ Q_2=C_2V \] where \(V\) is the battery voltage.

Step 2: Capacitance after insertion of dielectric When a dielectric of dielectric constant \(K\) is inserted completely between the plates, \[ C_1' = KC_1 \] and \[ C_2' = KC_2 \] Thus, both capacitances increase by a factor \(K\).

Step 3: Effect on charge stored Since the battery remains connected, \[ V=\text{constant} \] Using \[ Q=CV \] the new charge on capacitor \(C_1\) becomes \[ Q_1' = C_1'V \] \[ Q_1'=(KC_1)V \] \[ Q_1'=KQ_1 \] Similarly, for capacitor \(C_2\), \[ Q_2' = C_2'V \] \[ Q_2'=(KC_2)V \] \[ Q_2'=KQ_2 \] Hence, the charge stored on each capacitor increases \(K\) times. Result for part (i): \[ \boxed{Q_1'=KQ_1} \] \[ \boxed{Q_2'=KQ_2} \] Thus, the charge on each capacitor increases by a factor \(K\).

Step 4: Initial energy stored The energy stored in a capacitor is \[ U=\frac{1}{2}CV^2 \] Initially, \[ U_1=\frac{1}{2}C_1V^2 \] and \[ U_2=\frac{1}{2}C_2V^2 \]

Step 5: Energy stored after dielectric insertion Since voltage remains constant, \[ U_1'=\frac{1}{2}C_1'V^2 \] Substituting \(C_1'=KC_1\), \[ U_1'=\frac{1}{2}(KC_1)V^2 \] \[ U_1'=KU_1 \] Similarly, \[ U_2'=\frac{1}{2}C_2'V^2 \] \[ U_2'=\frac{1}{2}(KC_2)V^2 \] \[ U_2'=KU_2 \] Therefore, the energy stored in each capacitor also increases by a factor \(K\). Result for part (ii): \[ \boxed{U_1'=KU_1} \] \[ \boxed{U_2'=KU_2} \] Thus, the energy stored in each capacitor becomes \(K\) times its original value. Final Answer: When a dielectric slab of dielectric constant \(K\) is inserted into both capacitors while they remain connected to a battery:
• [(i)] The charge on each capacitor increases \(K\) times. \[ \boxed{Q'=KQ} \]
• [(ii)] The energy stored in each capacitor increases \(K\) times. \[ \boxed{U'=KU} \]