A parallel plate capacitor filled with a medium of dielectric constant 10 , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15 . Then the energy of capacitor will

Step-by-step solution

When a dielectric of dielectric constant K is inserted between the plates of a capacitor C 0 , the capacitance becomes K C 0 . Energy stored in a capacitor is given by, U = 1 2 C V 2 . Now, U i = 1 2 K 1 C 0 V 2 and U f = 1 2 K 2 C 0 V 2 ⇒ Δ U = U f - U i = 1 2 K 2 - K 1 C 0 V 2 Percentage change in the energy will be, Δ U U i × 100 = 1 2 × 5 × C 0 2 1 2 × 10 × C 0 2 × 100 = 50 %