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

We have a parallel plate capacitor maintained at a fixed voltage $$V$$ by a battery, and its dielectric constant is changed from $$K_1 = 10$$ to $$K_2 = 15$$.

Because the battery keeps the voltage constant, the energy stored in the capacitor is given by the well-known formula

$$U = \frac{1}{2} C V^2$$

Inserting a dielectric multiplies the vacuum capacitance $$C_0$$ by the dielectric constant $$K$$ so that $$C = K C_0$$. Hence the stored energy becomes

$$U = \frac{1}{2} K C_0 V^2$$

Comparing the energies before and after introducing the dielectric gives

$$\frac{U_2}{U_1} = \frac{K_2}{K_1} = \frac{15}{10} = 1.5$$

Therefore, the percentage increase is

$$\text{Percentage increase} = \left(\frac{U_2 - U_1}{U_1}\right)\times 100 = (1.5 - 1)\times 100 = 50\%$$

Thus the energy stored in the capacitor increases by 50%, corresponding to Option A.