Question

Two conducting spheres of radii $R_1$ and $R_2$ are charged with charges $Q_1$ and $Q_2$. If they are brought in contact, there is

• A. no increase in the energy of the system
• B. an increase in the energy of the system if $Q_1{R}_2\ne Q_2{R}_1$
• C. always a decrease in the energy of system
• D. an decrease in the energy of system if $Q_1{R}_2\ne Q_2{R}_1$

Answer 1

The correct option is D an decrease in the energy of system if $Q_1{R}_2\ne Q_2{R}_1$


From the definition of capacitance,

$Q=CV$

$\Rightarrow V=\frac{Q}{C}.......\left(1\right)$

Here, $C_1=4\pi {ϵ}_0{R}_1,C_2=4\pi {ϵ}_0{R}_2$

On bringing the conducting spheres in contact, the potential of the system of spheres will become equal.

$\therefore V_1=V_2$

$\Rightarrow \frac{Q_1}{4\pi {ϵ}_0{R}_1}=\frac{Q_2}{4\pi {ϵ}_0{R}_2}$

$\Rightarrow \frac{Q_1}{R_1}=\frac{Q_2}{R_2}$

Therefore, the condition for both sphere to have same electrostatic potential at initial condition is

$Q_1{R}_2=Q_2{R}_1$

If the above condition is fulfilled for the initial charges on the sphere, then there will be no flow of charges between spheres.

$\Rightarrow$ However $Q_1{R}_1\ne Q_2{R}_2,$ suggests that the charges on spheres pointing towards the unequal potential of spheres.

Therefore, if spheres are brought in contact, charge transfer will occur accompanied by loss of energy of system.

$\therefore$ option $\left(d\right)$ is the correct choice.

$\begin{array}{c}\text{Why this question?}\\ \text{Tip: Always remember that whenever}\\ \text{there is charge transfer, the fraction}\\ \text{of total potential energy is dissipated}\\ \text{in form of heat produced. Thus, final}\\ \text{energy of system decreases.}\end{array}$