Question

Two circular loops of radii $0.05\text{m}$ and $0.09\text{m}$ are arranged such that their axes coincide and their centres are $0.12\text{m}$ apart. Charge of ${10}^{-6}\text{C}$ is spread uniformly on each loop. The potential difference between the centres of the loops is

• A. $71\text{kV}$
• B. $64\text{kV}$
• C. $95\text{kV}$
• D. $82\text{kV}$

Answer 1

The correct option is A $71\text{kV}$

Given that,
Charge on each loop, $q={10}^{-6}\text{C}$
Radius of larger ring, $R_1=0.09\text{m}$
Radius of smaller ring, $R_2=0.05\text{m}$
Distance between centre of loops, $x=0.12\text{m}$

The electric potential of uniformly charged ring of radius $R$ at axial distance $x$ from the centre of ring is given by.
$V=\frac{kQ}{\sqrt{x^2+R^2}}$

The potential at the centre of a ring will be due to charge on both the rings as every element of the ring is at constant distance from the centre.

The electric potential at the centre of larger ring due to charge on larger ring is
$V_1=\frac{kq}{R_1}$
The electric potential at the centre of smaller ring due to charge on larger ring is
$V_1^{\prime }=\frac{kq}{\sqrt{x^2+R_1^2}}$

The electric potential at the centre of smaller ring due to charge on smaller ring is
$V_2=\frac{kq}{R_2}$
The electric potential at the centre of larger ring due to charge on smaller ring is
$V_2^{\prime }=\frac{kq}{\sqrt{x^2+R_2^2}}$

The net electric potential at the centre of larger ring is
$V_{net}=V_1+V_2^{\prime }$
The net electric potential at the centre of smaller ring is
$V_{net}^{\prime }=V_1^{\prime }+V_2$

The potential difference between the centre of the loops is
$\Delta V=V_{net}-V_{net}^{\prime }$
$\Rightarrow \Delta V=(V_1+V_2^{\prime })-(V_1^{\prime }+V_2)$
$\Rightarrow \Delta V=(\frac{kq}{R_1}+\frac{kq}{\sqrt{x^2+R_2^2}})-(\frac{kq}{\sqrt{x^2+R_1^2}}+\frac{kq}{R_2})$
$\Rightarrow \Delta V=kq\left(\right(\frac{1}{R_1}+\frac{1}{\sqrt{x^2+R_2^2}})-(\frac{1}{\sqrt{x^2+R_1^2}}+\frac{1}{R_2}\left)\right)$

$\Rightarrow \Delta V=9\times {10}^9\times {10}^{-6}\left(\right(\frac{1}{0.09}+\frac{1}{\sqrt{{0.12}^2+{0.05}^2}})-(\frac{1}{\sqrt{{0.12}^2+{0.09}^2}}+\frac{1}{0.05}\left)\right)$
$\Delta V=-70769\text{V}\approx -71\text{kV}$
Thus, the potential difference between the centre of the loops is $71\text{kV}$
Hence, option (b) is correct.