Question

Two spheres of radii $3$cm and $5$cm are charged to potentials $3000$V and $4500$V respectively. They are then connected by a thin metallic wires. The loss of electric energy in this process is?

• A. $2\times {10}^{-8}$J
• B. $4\times {10}^{-8}$J
• C. $62\times {10}^{-4}$J
• D. $2.36\times {10}^{-7}$J

Answer 1

The correct option is D $2.36\times {10}^{-7}$J

We know that the potential on a sphere is $\frac{KQ}{r}$

Let the charge on the first sphere be $Q_1$ and that on the second sphere be $Q_2$ and their radii being $r_1$ and $r_2$

Hence $\frac{K{Q}_1}{r_1}=3000$

and $\frac{K{Q}_2}{r_2}=4500$

On connecting these two spheres their potential will become equal

Putting the value of $K=9\times {10}^{9}N{m}^2{C}^{-2}$ and $r_1=3cm$ and $r_2=5cm$

$Q_1={10}^{-8}$

and $Q_2=2.5\times {10}^{-8}$

Now let us suppose the charges on the sphere after connecting them would be $q_1$ and $q_2$

Hence $\frac{K{q}_1}{r_1}=\frac{K{q}_2}{r_2}$

also $q_1+q_2=Q_1+Q_2$

Putting the values we get

$q_1=1.3125\times {10}^{-8}$ and $q_2=2.1875\times {10}^{-8}$

So initial energy, $E_1$= $\frac{3K{Q_1}^2}{5r_1}+\frac{3K{Q_2}^2}{5r_2}$

and after connecting the energy will be $E_2$= $\frac{3K{q_1}^2}{5r_1}+\frac{3K{q_2}^2}{5r_2}$

Hence the loss of energy will be $E_1-E_2=2.36\times {10}^{-7}J$