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Q. A capacitor has some dielectric between its plates, and the capacitor is connected to a dc source. The battery is now disconnected and then the dielectric is removed, then

Electrostatic Potential and Capacitance

Solution:

When the capacitor is connected to $dc$ source, it gets charged. The battery is then disconnected, so no more charge can flow in. On removing dielectric, capacitance decreases
Energy stored $\left(u=\frac{q^{2}}{2C}\right)$ will increase
Potential $\left(V=\frac{q}{C}\right)$ will also increase
Electric field $\left(E=\frac{V}{D}\right)$ will increase