Two identical capacitors 1 and 2 are connected in series to...

Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of constant $\mathrm{K} . \mathrm{Q}_{1}$ and $\mathrm{Q}_{2}$ are the charges stored in 1 and 2 . Now, the dielectric slab is removed and the corresponding charges are $Q_{1}^{\prime}$ and $Q_{2}^{\prime} 2$. Then (A) $\left(\mathrm{Q}_{1}^{1} / \mathrm{Q}_{1}^{1}\right)[(\mathrm{K}+1) / \mathrm{K}]$ (B) $\left(\mathrm{Q}_{1}^{1} / \mathrm{Q}_{1}^{1}\right)(\mathrm{K} / 2)$ (C) $\left(\mathrm{Q}_{2}{ }^{1} / \mathrm{Q}_{2}\right)[(\mathrm{K}+1) /(2 \mathrm{~K})]$ (D) $\left(\mathrm{Q}_{2}{ }^{1} / \mathrm{Q}_{2}\right)[(\mathrm{K}+1) / \mathrm{K}]$

Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of constant $\mathrm{K} . \mathrm{Q}_{1}$ and $\mathrm{Q}_{2}$ are the charges stored in 1 and 2 . Now, the dielectric slab is removed and the corresponding charges are $Q_{1}^{\prime}$ and $Q_{2}^{\prime} 2$. Then
(A) $\left(\mathrm{Q}_{1}^{1} / \mathrm{Q}_{1}^{1}\right)[(\mathrm{K}+1) / \mathrm{K}]$
(B) $\left(\mathrm{Q}_{1}^{1} / \mathrm{Q}_{1}^{1}\right)(\mathrm{K} / 2)$
(C) $\left(\mathrm{Q}_{2}{ }^{1} / \mathrm{Q}_{2}\right)[(\mathrm{K}+1) /(2 \mathrm{~K})]$
(D) $\left(\mathrm{Q}_{2}{ }^{1} / \mathrm{Q}_{2}\right)[(\mathrm{K}+1) / \mathrm{K}]$

Key Concepts

Relationship between Charge, Capacitance, and Voltage

The fundamental relation Q = CV underpins capacitor behavior where Q is the stored charge, C is the capacitance, and V is the voltage across the capacitor. Alterations in capacitance, such as those due to the insertion or removal of a dielectric, affect either the voltage or the charge depending on whether the circuit is connected to a constant voltage source or is isolated. Understanding this relation is crucial for analyzing how charges redistribute when circuit conditions change.

Effect of Removing a Dielectric

Removing the dielectric from a capacitor decreases its capacitance, because the dielectric is no longer enhancing the capacitor's ability to store charge. Depending on whether the capacitor remains connected to the battery or is isolated, the removal of the dielectric leads to different changes in voltage and stored energy. This change in capacitance directly results in a change in the charge distribution when comparing the states with and without the dielectric.

Dielectric Materials in Capacitors

A dielectric material, when inserted between the plates of a capacitor, increases its capacitance by a factor equal to the dielectric constant (K). The dielectric reduces the effective electric field within the capacitor, allowing it to store more charge for the same applied voltage. This concept is key in designing capacitors with desired storage properties and directly affects the relationship between charge, voltage, and stored energy.

Capacitors in Series

In a series circuit, the same charge appears on each capacitor, even though the voltage across each capacitor may differ. The overall or equivalent capacitance is lower than any individual capacitance, calculated by the reciprocal of the sum of reciprocals of the individual capacitances. This principle is essential in understanding how the voltage divides among capacitors based on their individual capacitances.

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