Question

A capacitor of unknown capacitance is connected across a battery of $$V$$ volts. The charge stored in it is $$360\mu C$$. When potential across the capacitor is reduced by $$120V$$, the charge stored in it becomes $$120\mu C$$.

Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by$$120V$$?

Answer 1

Hint: This numerical problem can be solved by using the fact that value of the capacitance is actually determined only by the physical characteristics of the capacitor. This is because according to the principle of a capacitor, its capacitance increases only when the voltage decreases. We can use this expression to solve this numerical problem.

Formula Used:
We know that capacitance is actually the ratio of charge and potential. This can be mathematically represented by the mathematical expression given below:
$$C=Q/V$$
We will use the above mathematical expression to solve the given numerical problem.

Complete step by step answer:
First we will have to consider the initial voltage of the capacitor to be $$V_1=V$$. Now, the initial charge of the capacitor is given in the numerical problem as $$q_1=360\mu C$$.
Now, using these values in the expression for capacitance of the conductor, we get:
$C={\displaystyle \frac{q_1}{V_1}}={\displaystyle \frac{360}{V}}$
Now, according to the question, the voltage is reduced by $$120V$$ thus, $V_2=V-120$. The charge on the capacitor is now equal to $$120\mu C$$.
Now, we will again find the capacitance by using the formula given above:
Thus, $C={\displaystyle \frac{q_2}{V_2}}={\displaystyle \frac{120}{(V-120)}}$
Now, we can say that the capacitance of the capacitor remains constant as there is no change in area of the plates, medium and distance between plates. Thus, we can equate the value of the capacitance from both the equations. Thus,
${\displaystyle \frac{360}{V}}={\displaystyle \frac{120}{(V-120)}}$
Using $V$ as the subject of the formula, we get:
$\Rightarrow 3(V-120)=V$
Solving for $V$ we have:
$\Rightarrow V={\displaystyle \frac{360}{2}}=180V$
Now, we can use this value of the potential to obtain the unknown capacitance by substituting this in the above formula:
$C={\displaystyle \frac{130}{180}}=2\mu F$
Thus, the answer to (i) is that the potential is equal to $180V$ and the unknown capacitance is equal to $2\mu F$.
Now, it is given that the voltage applied is increased by $120V$. Thus, we have to add this increase in voltage to the original potential. Thus,
New voltage, $V^{\prime }=180+120=300V$
Now, the charge stored in a capacitor is given by the mathematical formula $$q^{\prime }=C{V}^{\prime }$$. Thus, substituting these values in the equation, we get:
$q^{\prime }=2\times 300=600\mu C$
Thus the answer for (ii) is equal to $600\mu F$.

Note: It is important to proceed in a step by step manner while solving numerical problems related to capacitance. Students usually get confused while working on such numerical problems as there are many increases and decreases in currents and voltages which might confuse them.