Question 490: Electrostats & Capacitance

Question

Two identical capacitors A and B are connected in series to a battery of E.M.F., 'E'. Capacitor B contains a slab of dielectric constant $\mathrm{K} . \mathrm{Q}_{\mathrm{A}}$ and $\mathrm{Q}_{\mathrm{B}}$ are the charges stored in A and B . When the dielectric slab is removed, the corresponding charges are $\mathrm{Q}_{\mathrm{A}}^{\prime}$ and $\mathrm{Q}_{\mathrm{B}}^{\prime}$. Then

Accepted Answer

When two capacitors, A and B, are connected in series, they will have the same charge initially due to the series configuration. Consider the capacitors connected to a battery of voltage $E$.
Initial Configuration with Dielectric in Capacitor B:
The capacitance of capacitor B with a dielectric is $C_B' = KC$, where $C$ is the capacitance of capacitor A, and $K$ is the dielectric constant.
For capacitors in series,
$$ \frac{1}{C_{	ext{eq}}} = \frac{1}{C} + \frac{1}{KC} = \frac{K + 1}{KC}. $$
$$ C_{	ext{eq}} = \frac{KC}{K + 1}. $$
The charge on the capacitors in series is given by:
$$ Q = C_{	ext{eq}} \cdot E = \frac{KC}{K + 1} \cdot E. $$
Charges on Each Capacitor:
Since $Q_A = Q_B = Q$ (initially when the dielectric is present in B),
$$ Q_A = Q_B = \frac{KC}{K + 1} \cdot E. $$
After Removing the Dielectric from Capacitor B:
Without the dielectric, the capacitance of B becomes $C$. The equivalent capacitance is now:
$$ \frac{1}{C_{	ext{eq}}'} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}. $$
$$ C_{	ext{eq}}' = \frac{C}{2}. $$
The charge on the new configuration is:
$$ Q' = C_{	ext{eq}}' \cdot E = \frac{C}{2} \cdot E. $$
Since the capacitors are still in series, the charges on A and B are equal:
$$ Q_A' = Q_B' = Q'. $$
Comparing the Charges:
For capacitor A:
$$ \frac{Q_A'}{Q_A} = \frac{Q'}{Q_A} = \frac{\frac{C}{2} \cdot E}{\frac{KC}{K + 1} \cdot E} = \frac{K + 1}{2K}. $$
For capacitor B:
$$ \frac{Q_B'}{Q_B} = \frac{\frac{C}{2} \cdot E}{\frac{KC}{K + 1} \cdot E} = \frac{K + 1}{2K}. $$
The correct option in this context would be:
Option D:
$$ \frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}} $$

Answer

$\mathrm{\frac{Q_A^{\prime}}{Q_A}=\frac{K}{2}}$

Answer

$\frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2}$

Answer

$\frac{\mathrm{Q}_{\mathrm{A}}^{\prime}}{\mathrm{Q}_{\mathrm{A}}}=\frac{\mathrm{K}+1}{\mathrm{~K}}$

Answer

$\frac{\mathrm{Q}_{\mathrm{B}}^{\prime}}{\mathrm{Q}_{\mathrm{B}}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}}$

Electrostats & Capacitance

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