s. A dielectric slab is slowly inserted between the plates of a parallel plate c

Question

s. A dielectric slab is slowly inserted between the plates of a parallel plate capacitor while (A) The capacitance, the potential difference between the plates, and the charge on the (B) The capacitance, the potential difference between the plates, and the charge on the the capacitor is connected to a battery. As it is being inserted positive plate all increase. positive plate all decrease. decreases, and the capacitance remains the same. difference between the plates remains the same. between the plates remains the same. (C) The potential difference between the plates increases, the charge on the positive plate (D) The capacitance and the charge on the positive plate decrease, but the potential (E) The capacitance and the charge on the plate increase, but the potential difference

Explanation / Answer

Option D is the correct answer.

When capacitor is being connected to the battery. Potential diffrence across the capacitor reamins constant.
And that value is equal to potential diffrence across the terminals of the battery.

But capacitance of the capacitor increases. so, charge on the capacitor also increases.

Before insering dielectric.

potential difference across capacitor = V

Capacitnace = C

charge = Q

After insering dielectric.

potential difference across capacitor = V

Capacitnace = k*C

charge = k*Q

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