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

The correct option is A Increase
$$C=\cfrac QV$$   $$\therefore V=\cfrac QC$$
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.

Physics
NCERT
Standard XII

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