A parallel plate capacitor is charged by a battery after charging the capacitor, the battery is disconnected and if a dielectric plate is inserted between the plates, then which one of the following statements is not correct

Question

A parallel plate capacitor is charged by a battery after charging the capacitor, the battery is disconnected and if a dielectric plate is inserted between the plates, then which one of the following statements is not correct (A) increase in the stored energy (B) decrease in the potential difference (C) decrease in the electric field (D) increase in the capacitance

Tardigrade Admin · Accepted Answer

New capacitance, C=KC0 New electric field, E=(E0/K) New potential difference Δ V=(Q0/C)=(Q0/K C0) Δ V=(Δ V0/K)

Q. A parallel plate capacitor is charged by a battery after charging the capacitor, the battery is disconnected and if a dielectric plate is inserted between the plates, then which one of the following statements is not correct

increase in the stored energy

decrease in the potential difference

decrease in the electric field

increase in the capacitance

New capacitance,
$C = K C_{0}$
New electric field,
$E = \frac{E _{0}}{K}$
New potential difference
$Δ V = \frac{Q _{0}}{C} = \frac{Q _{0}}{K C _{0}}$
$Δ V = \frac{Δ V _{0}}{K}$

Q. A parallel plate capacitor is charged by a battery after charging the capacitor, the battery is disconnected and if a dielectric plate is inserted between the plates, then which one of the following statements is not correct

increase in the stored energy

decrease in the potential difference

decrease in the electric field

increase in the capacitance

New capacitance, C=KC0​ New electric field, E=KE0​​ New potential difference ΔV=CQ0​​=KC0​Q0​​ ΔV=KΔV0​​

New capacitance,
$C = K C_{0}$
New electric field,
$E = \frac{E _{0}}{K}$
New potential difference
$Δ V = \frac{Q _{0}}{C} = \frac{Q _{0}}{K C _{0}}$
$Δ V = \frac{Δ V _{0}}{K}$