Question

Two capacitors ${C_1}$ and ${C_2}$ are charged to $120V$ and $200V$ respectively. It is found that by connecting them together, the potential on each one can be made zero. Then:
A) $3{{\text{C}}_{\text{1}}} + 5{{\text{C}}_2} = 0$
B) $5{{\text{C}}_1} = 3{{\text{C}}_2}$
C) ${\text{9}}{{\text{C}}_1} = 4{{\text{C}}_2}$
D) ${\text{3}}{{\text{C}}_1} = 5{{\text{C}}_2}$

Answer 1

Capacitor: A capacitor is an Electrical device that stores electrical energy in an electric field. It is a two-terminal device. The effect of the capacitor is called capacitance. It is calculated in Farad $\left( F \right)$.
Potential difference: It is defined as the difference in the energy that the charge carrier has between the two points in a circuit.
When the resistance of the wires is much smaller than the resistance of the other elements in the circuit, the potential difference drops down to zero.

Formula used:
The relation between charge, capacitance and voltage of the circuit ${\text{Q = CV}}$, $Q =$ charge in the capacitor, $C =$ capacitance of the capacitor, $V =$ potential difference in the circuit.

Complete step-by-step solution:
It is given details:
Capacitance of the two capacitors is given as $\;{C_1}$ and ${C_2}$. ${{\text{V}}_1} = 120{\text{V}}$, ${{\text{V}}_2} = 200{\text{V}}$
The potential difference in the given set will be zero if and only if the charge in the two capacitors $\;{C_1}$ and $\;{C_2}$ is equal. To find the relation between the two capacitance capacitors $\;{C_1}$ and $\;{C_2}$.
That is, ${{\text{Q}}_1} = {{\text{Q}}_2}$

From the above equation of charge, we can say that, ${{\text{C}}_1}{{\text{V}}_1} = {{\text{C}}_2}{{\text{V}}_2}$
Substituting the values, we get that, ${\text{120}}{{\text{C}}_1} = 200{{\text{C}}_2}$
$\Rightarrow 3{{\text{C}}_1} = 5{{\text{C}}_2}$
This is the required relationship between the capacitors ${C_1}$ and ${C_2}$

Hence the correct option is (D).

Note:
1. The capacitance of a capacitor increases with the decrease in the distance between the plates.
2. The material inserted between the plates of a capacitor also changes the capacitance of the capacitor.
3. The effective increase in the area of the plates in the capacitor decreases the potential difference between the plates and increases the capacitance of the capacitor.