When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

Question

When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge (A) is less (B) is same (C) is more (D) may be less or more depending on the nature of the material introduced

Tardigrade Admin · Accepted Answer

When dielectric is introduced, the capacitance will increase and as the battery remains connected, so the voltage will remain constant. Hence according to Q=C V, the charge will increase.

Q. When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

is less

is same

is more

may be less or more depending on the nature of the material introduced

When dielectric is introduced, the capacitance will increase and as the battery remains connected, so the voltage will remain constant. Hence according to $Q = C V$ , the charge will increase.

Q. When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

is less

is same

is more

may be less or more depending on the nature of the material introduced

When dielectric is introduced, the capacitance will increase and as the battery remains connected, so the voltage will remain constant. Hence according to Q=CV, the charge will increase.

When dielectric is introduced, the capacitance will increase and as the battery remains connected, so the voltage will remain constant. Hence according to $Q = C V$ , the charge will increase.