A parallel-plate capacitor of capacitance is charged to a potential and then…

PHYSICS · Single correct · CBSE · Physics · Electrostatic Potential and Capacitance · Electrostatic Potential and Capacitance

Problem

A parallel-plate capacitor of capacitance

C_0

is charged to a potential

V_0

and then disconnected from the battery. A dielectric slab of constant

\kappa

is completely inserted between the plates. The new potential difference between the plates is (a)

V_0

(b)

\kappa V_0

(c)

\dfrac{V_0}{\kappa}

(d)

\dfrac{V_0}{\kappa^{2}}