Question

A capacitor of capacitance $10\mu F$ is charged a potential 50V with a battery. The battery is now disconnected and an additional charge $200\mu C$ is given to the positive plate of the capacitor. The potential difference across the capacitor will be:
A. 50V
B. 80V
C. 100V
D. 60V

Answer 1

Hint: We know that the relation between charge and capacitor. The amount of charge that moves into the plates depends upon the capacitance and the applied voltage. Capacitance is a function of the projected area of one plate to the other and the dielectric constant of the medium between the two plates.

Complete step by step answer:
Charged acquired by the plates of the capacitor
${{\text{q}}_0} = {\text{CV = 10}}\mu {\text{F}} \times {\text{50V = 500}}\mu {\text{C}}$
Where $q_0$ is a charge, C is capacitance; V is voltage (potential).
When an additional charge $200\mu {\text{C}}$ is given to the positive plate, it distributes both sides of the plate equally. So charge in inner side of each plate will be
i.e. total charge on positive plate becomes $500\mu {\text{C + 100}}\mu {\text{C = 600}}\mu {\text{C}}$
Now the potential across the capacitor will be $ = \dfrac{{600\mu {\text{c}}}}{{10\mu {\text{F}}}} = 60V$

$\therefore$ The potential across the capacitor is 60V. So, Option (D) is correct.

Additional information:
If the wires to the battery are disconnected, the charge remains on the plates and the voltage across the plates remains the same. If the wires are connected to each other, the current will flow and the capacitor will discharge. Then there will be no voltage across the capacitor or any charge on the plates.

Note:
A capacitor will store energy when an electric charge is forced onto its plates from a power source. A capacitor will still retain this charge even after disconnection from the power source.