Question

A capacitor of capacitance C is connected to battery of emf $V_0$. Without removing the battery, a dielectric of strength ${\epsilon }_r$ is inserted between the parallel plates of the capacitor C, then the charge on the capacitor is :

• A. $C{V}_0$
• B. ${\epsilon }_{r}C{V}_0$
• C. $\frac{C{V}_0}{{\epsilon }_r}$
• D. none of these

Answer 1

The correct option is B ${\epsilon }_{r}C{V}_0$

As the battery is not removed so the potential remains constant and it is equal to $V_0$.
Due to insert the dielectric the capacitance $C$ will become $C^{\prime }={\epsilon }_{r}C$
Thus the charge on the capacitor $Q=C^{\prime }{V}_0={\epsilon }_{r}C{V}_0$