  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  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.PhysicsNCERTStandard XII

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