An infinitely long thin wire, having a uniform charge density per unit length of 5nC/m , is passing through a spherical shell of radius 1m , as shown in the figure. A 10nC charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points P and R, in Volt, is

$[\text{Given:}\text{In SI units}\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9,\ln 2=0.7.\text{Ignore the area pierced by the wire.}]$

The correct answer is 171.

1. Understanding the Given Configuration and Data:

From the question and the provided figure, we have:

• Radius of the spherical shell, $R=1\text{m}$.

• Uniform charge on the spherical shell, $Q=10\text{nC}=10\times {10}^{-9}\text{C}$.

• Linear charge density of the infinitely long wire, $\lambda =5\text{nC/m}=5\times {10}^{-9}\text{C/m}$.

• Point $P$ is located inside the spherical shell at a perpendicular distance $r_P=0.5\text{m}$ from the wire.

• Point $R$ is located outside the spherical shell at a perpendicular distance $r_R=2\text{m}$ from the wire.

• The line connecting the center of the shell to points $P$ and $R$ is perpendicular to the wire, passing through the center of the sphere.

• Given values: $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\text{N m}{}^2\text{/C}{}^2$ and $\ln 2=0.7$.

2. Superposition Principle for Potential:

By the principle of superposition, the net potential at any point is the sum of the potential due to the spherical shell and the potential due to the infinite wire:

$V=V_{\text{shell}}+V_{\text{wire}}$

3. Electric Potential due to the Spherical Shell:

• Since point $P$ is inside the spherical shell ($r_{\text{shell, P}}=0.5\text{m}<R$), the potential inside a uniformly charged thin spherical shell is constant and equal to the potential on its surface:

$V_{\text{shell}}\left(P\right)=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$

• Point $R$ is outside the spherical shell ($r_{\text{shell, R}}=2\text{m}>R$), so the potential outside is given by:

$V_{\text{shell}}\left(R\right)=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r_{\text{shell, R}}}$

Thus, the potential difference contributed by the shell is:

$\Delta V_{\text{shell}}=V_{\text{shell}}\left(P\right)-V_{\text{shell}}\left(R\right)=\frac{1}{4\pi {\epsilon }_0}Q(\frac{1}{R}-\frac{1}{r_{\text{shell, R}}})$

Substituting the numerical values:

$\Delta V_{\text{shell}}=(9\times {10}^9)\times (10\times {10}^{-9})\times (\frac{1}{1}-\frac{1}{2})=90\times 0.5=45\text{V}$

4. Electric Potential Difference due to the Infinite Wire:

The potential difference between two points at perpendicular distances $r_P$ and $r_R$ from an infinitely long wire of uniform charge density $\lambda$ is given by:

$\Delta V_{\text{wire}}=V_{\text{wire}}\left(P\right)-V_{\text{wire}}\left(R\right)=\frac{\lambda }{2\pi {\epsilon }_0}\ln \left(\frac{r_R}{r_P}\right)$

We can rewrite $\frac{1}{2\pi {\epsilon }_0}=2\times \frac{1}{4\pi {\epsilon }_0}=2\times (9\times {10}^9)=18\times {10}^9\text{N m}{}^2\text{/C}{}^2$.

Substituting the given values into the equation:

$\Delta V_{\text{wire}}=(18\times {10}^9)\times (5\times {10}^{-9})\times \ln \left(\frac{2}{0.5}\right)$

$\Delta V_{\text{wire}}=90\times \ln \left(4\right)=90\times \ln \left(2^2\right)=90\times 2\ln 2$

Using $\ln 2=0.7$:

$\Delta V_{\text{wire}}=90\times 2\times 0.7=90\times 1.4=126\text{V}$

5. Calculating Total Potential Difference:

The magnitude of the potential difference between points $P$ and $R$ is:

$|V_P-V_R|=\Delta V_{\text{shell}}+\Delta V_{\text{wire}}=45\text{V}+126\text{V}=171\text{V}$