An infinitely long thin wire, having a uniform charge density per unit length of 5 nC/m, is passing through a spherical shell of radius 1 m, as shown in the figure. A 10 nC 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 ______.

[Given: In SI units $\frac{1}{4\mathrm{\pi <!--\; \pi \; -->}{\mathrm{ϵ<!--\; ϵ\; -->}}_0}=9\times <!--\; \times \; -->{10}^9,\mathrm{ln }<!--\; \; -->2=0.7$. Ignore the area pierced by the wire.]

The correct answer is 171.

1. Analysis of the Image and Given Parameters:
Based on the provided diagram:

• An infinitely long thin wire runs vertically through the center of a spherical shell of radius $R_s=1\text{m}$.
• The linear charge density of the wire is $\lambda =5\text{nC/m}=5\times {10}^{-9}\text{C/m}$.
• The spherical shell carries a uniformly distributed charge of $Q=10\text{nC}=10\times {10}^{-9}\text{C}$.
• Point P lies at a perpendicular distance of $r_P=0.5\text{m}$ from the wire, which is inside the spherical shell since $r_P<R_s$.
• Point R lies at a perpendicular distance of $r_R=2\text{m}$ from the wire, which is outside the spherical shell since $r_R>R_s$.

By the principle of superposition, the total electric potential at any point is the sum of the potentials due to the spherical shell and the infinite wire:
$V=V_{\text{shell}}+V_{\text{wire}}$

2. Potential Difference due to the Spherical Shell:
For a uniformly charged spherical shell:

• Inside the shell ($r<R_s$), the potential is constant and equal to its value at the surface:

$V_{\text{shell}}(P)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R_s}$
Substituting the given values:
$V_{\text{shell}}(P)=(9\times {10}^9)\times \frac{10\times {10}^{-9}}{1}=90\text{V}$

• Outside the shell ($r>R_s$), the potential is equivalent to that of a point charge located at the center:

$V_{\text{shell}}(R)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r_R}$
Substituting the given values:
$V_{\text{shell}}(R)=(9\times {10}^9)\times \frac{10\times {10}^{-9}}{2}=45\text{V}$
Thus, the contribution of the shell to the potential difference between points P and R is:
$\Delta V_{\text{shell}}=V_{\text{shell}}(P)-V_{\text{shell}}(R)=90-45=45\text{V}$

3. Potential Difference due to the Infinitely Long Wire:
The electric field at a distance $r$ from an infinitely long wire with linear charge density $\lambda$ is:
$E(r)=\frac{\lambda }{2\pi {ϵ}_0}\frac{1}{r}$
The potential difference between points P and R due to the wire is:
$V_{\text{wire}}(P)-V_{\text{wire}}(R)={\int }_{r_P}^{r_R}E(r)dr=\frac{\lambda }{2\pi {ϵ}_0}\text{ln}\left(\frac{r_R}{r_P}\right)$
Using the relation $\frac{1}{2\pi {ϵ}_0}=2\times \frac{1}{4\pi {ϵ}_0}=18\times {10}^9{\text{N m}}^2{\text{/C}}^2$:
$V_{\text{wire}}(P)-V_{\text{wire}}(R)=(18\times {10}^9)\times (5\times {10}^{-9})\times \text{ln}\left(\frac{2}{0.5}\right)$
$V_{\text{wire}}(P)-V_{\text{wire}}(R)=90\times \text{ln}(4)=90\times 2\text{ln}(2)=180\text{ln}(2)$
Given that $\text{ln}(2)=0.7$:
$V_{\text{wire}}(P)-V_{\text{wire}}(R)=180\times 0.7=126\text{V}$

4. Total Potential Difference:
The total potential difference between points P and R is:
$V(P)-V(R)=[V_{\text{wire}}(P)-V_{\text{wire}}(R)]+[V_{\text{shell}}(P)-V_{\text{shell}}(R)]$
$V(P)-V(R)=126+45=171\text{V}$
Therefore, the magnitude of the potential difference between points P and R is 171 V.