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) ${\text{ }}3{C_1}{\text{ }} = 5{C_2}$
(B) $3{C_1} + 5{\text{ }}{C_2} = 0$
(C) $9{C_1} = 4{C_2}$
(D) $5{C_1} = {\text{ }}3{C_2}$

Answer 1

First, find the value of charges of each capacitor. The value of the charge is the product of the capacitance and the potential difference of a capacitor.
Note that, we can find the potential difference across the two sides when the charges are unequal. If there is no difference in the value of the charges, the potential difference will be zero.

Formula used:
Charge of the first capacitor ${Q_1} = {C_1}{V_1}$
Charge of the second capacitor ${Q_2} = {C_2}{V_2}$
${C_1}$ and ${C_2}$ are the capacitances.

Complete step by step answer:
Let, the two conductors of capacitances ${C_1}$ and ${C_2}$ are charged to voltages ${V_1}$ and ${V_2}$ respectively.
Given that, ${V_1} = 120V$ and ${V_2} = 200V$
It is found that after connecting them through a wire, the potential difference becomes zero.
The potential on each other can be made zero only when the charges of the conductors are equal i.e
Charge of the first capacitor ${Q_1}$ = Charge of the second capacitor ${Q_2}$
${Q_1} = {C_1}{V_1}$ and ${Q_2} = {C_2}{V_2}$
$\therefore {C_1}{V_1} = {C_2}{V_2}$
$\Rightarrow 120{C_1} = 200{C_2}$
$\Rightarrow 3{C_1} = 5{C_2}$
So, the relation between the capacitance is $\Rightarrow 3{C_1} = 5{C_2}$.

Hence, the correct answer is option (A).

Note: The capacitance is the required charge to raise the potential to $1^\circ$ . it is a scalar quantity.
The capacitance of a conductor is defined by, $C = \dfrac{Q}{V}$
Where, $Q$ is the charge and $V$ is the raised potential or can say the potential difference.
Now if we connect two different conductors of charges ${Q_1}$ and ${Q_2}$ and capacitance ${C_1}$ and ${C_2}$ respectively, due to connection with each other the potential will be the same for both, this is called the common potential.
The common potential, $V = \dfrac{{{Q_1}}}{{{C_1}}} = \dfrac{{{Q_2}}}{{{C_2}}} = \dfrac{{{Q_1} + {Q_2}}}{{{C_1} + {C_2}}}$.