Question

Two identical capacitor $C_1$ and $C_2$ are connected in series with a battery. They are fully charged. Now dielectric slab is inserted between the plates of $C_2$. The potential difference across $C_1$ will

• A. Increase
• B. Decrease
• C. Remain same
• D. Depend on internal resistance of the cell

Answer 1

The correct option is A Increase

$C=\frac{Q}{V}$ $\therefore V=\frac{Q}{C}$

Whenever capacitors are connected in series, the charge accumulated in their plates is common. Hence $Q$ is common.

Initially, the capacitance of both is the same. $\therefore V$is same for both.

Whenever a dielectric material is inserted in between the plates of a capacitor, its capacitance increases. $Now\therefore C_2>C_1$

If we look at the above equations, we can note that for $Q$ to remain common for both, and for $C_2$ to increase, $V_2$ has to decrease.

$V=V_1+V_2$ here $V$ is the PD of the cell.

Since that doesn't change, and if $V_2$ decreases, $V_1$ has to increase.

The potential difference across $C_1$ will increase.