A parallel plate capacitor is made of two square parallel plates of area $A$ , and separated by a distance $d < < \sqrt A $ . The capacitor is connected to a battery with potential $V$ and allowed to fully charge. The battery is then disconnected. A square metal conducting slab also with area $A$ but thickness $\frac {d}{2}$ is then fully inserted between the plates, so that it is always parallel to the plates. How much work has been done on the metal slab by external agent while it is being inserted?
$ + \frac{1}{4}\,\frac{{{ \in _0}A}}{d}{V^2}$
$ - \frac{1}{2}\,\frac{{{ \in _0}A}}{d}{V^2}$
$ + \frac{1}{2}\,\frac{{{ \in _0}A}}{d}{V^2}$
$ - \frac{1}{4}\,\frac{{{ \in _0}A}}{d}{V^2}$
A $4 \;\mu\, F$ capacitor is charged by a $200\; V$ supply. It is then disconnected from the supply, and is connected to another uncharged $2 \;\mu\, F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Two identical capacitors have same capacitance $C$. One of them is charged to the potential $\mathrm{V}$ and other to the potential $2 \mathrm{~V}$. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :
If an electron enters into a space between the plates of a parallel plate capacitor at an angle $\alpha $ with the plates and leaves at an angle $\beta $ to the plates, the ratio of its kinetic energy while entering the capacitor to that while leaving will be
A parallel plate capacitor is charged fully by using a battery. Then, without disconnecting the battery, the plates are moved further apart. Then,
A $600\,pF$ capacitor is charged by $200\,V$ supply. It is then disconnected from the supply and is connected to another uncharged $600\,pF$ capacitor. Electrostatic energy lost in the process is $.........\,\mu J$.