Courses for Kids
Free study material
Offline Centres
Store Icon

Derive the formula for loss in energy on joining of two charged capacitors by a wire.

Last updated date: 14th Jun 2024
Total views: 392.7k
Views today: 5.92k
392.7k+ views
Hint: The given problem can be solved by using the concept of the charging and discharging of the capacitors. Capacitors are devices that store the electrical energy in their electric field. This energy depends on the potential difference across the plates of the capacitor and the capacitance of the capacitor.

Complete step by step solution:
Let the two capacitors be ${C_1}$ and ${C_2}$ at some potential differences ${V_1}$ and ${V_2}$ respectively. We are given that both the conductors are joined by a wire. As soon as they are connected by the wire, charge starts to flow from higher potential to lower potential. This flow of charge continues till they reach a common potential difference.
The common potential difference will be:
${V_{common}} = \dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$
Here, {V_{common}} is the common potential difference.
The initial energy stored in each capacitor is:
${E_1} = \dfrac{1}{2}{C_1}{V_1}^2 and {E_2} = \dfrac{1}{2}{C_2}{V_2}^2$
Here, ${E_1}$ and ${E_2}$ are the initial energy stored in both the capacitors.
After joining the wire, both the capacitors will reach a point of common potential difference, therefore the final energy in both the capacitors will be given as
${E_{final}} = \dfrac{1}{2}{C_1}{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2} + \dfrac{1}{2}{C_2}{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
$ \Rightarrow {E_{final}} = \dfrac{1}{2}\left( {{C_1} + {C_2}} \right){\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
Here, ${E_{final}}$ is the final energy stored in the capacitors.
The loss in energy will be given by the difference between initial energy and final energy;
$\left( {{E_1} + {E_2}} \right) - {E_{final}} = \left( {\dfrac{1}{2}{C_1}{V_1}^2 + \dfrac{1}{2}{C_2}{V_2}^2} \right) - \dfrac{1}{2}\left( {{C_1} + {C_2}} \right){\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)^2}$
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{2}\left( {{C_1}{V_1}^2 + {C_2}{V_2}^2} \right) - \left[ {\left( {{C_1} + {C_2}} \right){{\left( {\dfrac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}} \right)}^2}} \right]$
On solving the equation, we have
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}\left[ {\left( {{C_1}^2{V_1}^2 + {C_2}{C_1}{V_2}^2 + {C_2}{C_1}{V_1}^2 + {C_2}^2{V_2}^2} \right) - \left( {{C_1}^2{V_1}^2 + {C_2}{V_2}^2 + 2{C_1}{C_2}{V_1}{V_2}} \right)} \right]$
 $\Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}\left( {{C_1}{C_2}{V_2}^2 + {C_1}{C_2}{V_1}^2 - 2{C_1}{C_2}{V_1}{V_2}} \right)$
$ \Rightarrow \left( {{E_1} + {E_2}} \right) - {E_{final}} = \dfrac{1}{{2\left( {{C_1} + {C_2}} \right)}}{C_1}{C_2}{\left( {{V_1} - {V_2}} \right)^2}$
This is the required formula for loss of energy on joining of two charged conductors by a wire.

Note: In the derived formula, the capacitances cannot be negative individually. Also, the value of ${\left( {{V_1} - {V_2}} \right)^2}$ will be non-negative as it is a square value, this implies that the final value will be positive. Thus, initial energy was greater than final energy. Also, for ${V_1}$ = ${V_2}$ , the difference in energy will be zero. This means that no loss of energy takes place as there is no potential difference initially, there will be no flow of current.