Question

A capacitor of capacitance $C$ is charged to a potential difference $V_{0} .$ The terminals of the charged capacitor are then connected to those of an uncharged capacitor of capacitance $\frac{C}{2} .$

(a) Compute the original charge of the system.

(b) Find the final potential difference across each capacitor.

(c) Find the final energy of the system.

(d) Calculate the decrease in energy when the capacitors are connected.

(e) Where did the "lost" energy go?

Accepted Answer

(a):

Given that the original capacitance of the system is $C_{1} = C$ and potential difference across the capacitor is $V_{0}$ .

The charge on the original system before an uncharged capacitor was connected to it $Q = C V_{0}$ .

(b):

When the uncharge capacitor with capacitance $C_{2} = \frac{C}{2}$ is connected, the charge will flow from the charge capacitor to the uncharged capacitor.

The total charge across the two capacitor should be equal to the original charge.

$Q_{1} + Q_{2} = Q . . . (i)$

The potential difference across both the capacitors is same. So, $V = \frac{Q_{1}}{C_{1}} = \frac{Q_{2}}{C_{2}}$ .

$\Rightarrow \frac{Q_{1}}{C} = \frac{Q_{2}}{(\frac{C}{2})} \Rightarrow Q_{1} = 2 Q_{2} . . . (ii)$

Using equation $(ii)$ in $(i)$ , we get,

$Q_{2} = \frac{Q}{3}$ and $Q_{1} = \frac{2 Q}{3}$ .

Therefore, the potential difference is $V = \frac{Q}{3} \times \frac{2}{C} = \frac{2 Q}{3 C}$

$\Rightarrow V = \frac{2}{3} V_{0}$ .

(c):

Final energy of the system is $U_{f} = \frac{1}{2} \frac{{Q_{1}}^{2}}{C_{1}} + \frac{1}{2} \frac{{Q_{2}}^{2}}{C_{2}} . . . (iii)$

Using all the value in equation $(iii)$ , we get,

$U_{f} = \frac{1}{2} {(\frac{2 Q}{3})}^{2} \frac{1}{C} + \frac{1}{2} {(\frac{Q}{3})}^{2} \frac{2}{C}$

$\Rightarrow U_{f} = \frac{Q^{2}}{3 C} = \frac{1}{3} C {V_{0}}^{2} . . . (iv)$

(d):

Initial energy of the system is $U_{i} = \frac{1}{2} C {V_{0}}^{2}$ .

Final energy of the system is $U_{f} = \frac{1}{3} C {V_{0}}^{2}$ .

Decrease in energy, $∆ U = U_{f} - U_{i} = \frac{1}{3} C {V_{0}}^{2} - \frac{1}{2} C {V_{0}}^{2}$

$\Rightarrow ∆ U = - \frac{1}{6} C {V_{0}}^{2}$

(e):

The loss of energy is due to the heat in the wire, electromagnetic radiations, etc.

B M Sharma Solutions for Chapter: Capacitor and Capacitance, Exercise 1: Concept Application Exercise

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Capacitor and Capacitance, Exercise 1: Concept Application Exercise

Questions from B M Sharma Solutions for Chapter: Capacitor and Capacitance, Exercise 1: Concept Application Exercise with Hints & Solutions

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