Capacitors: Practice Problem & Solution
The energy and capacity of a charged parallel plate capacitor are $E$ and $C$ respectively. Now a dielectric slab of $\epsilon_{r}=6$ is inserted in it then energy and capacity becomes (Assuming charge on plates remains constant): (1999)
Solution Explained:
To solve this problem, we apply the core principles of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:
When charge $Q$ is constant, the new capacity is $C' = KC = 6C$. The new energy is $E' = \frac{Q^{2}}{2C'} = \frac{Q^{2}}{2(6C)} = \frac{E}{6}$.
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