Potential Difference Between Points Near Charged Wire and Spherical Shell

To find the magnitude of the potential difference between points $P$ and $R$, we can use the principle of superposition by considering the contributions from the uniformly charged spherical shell and the infinitely long thin wire separately.

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1. Contribution from the Spherical Shell

Let $P$ be the center of the spherical shell with radius $R_{\text{shell}} = 1\text{ m}$ and total charge $Q = 10\text{ nC} = 10 \times 10^{-9}\text{ C}$.

• At Point $P$ (Center of the Shell):
  Since point $P$ lies inside the spherical shell, the potential due to the shell is uniform and equal to its surface potential: $V_{\text{shell}}(P) = \frac{1}{4\pi\epsilon_0} \frac{Q}{R_{\text{shell}}}$ Substituting the given values: $V_{\text{shell}}(P) = (9 \times 10^9) \times \frac{10 \times 10^{-9}}{1} = 90\text{ V}$

• At Point $R$ (Outside the Shell):
  Point $R$ is located at a distance of $2\text{ m}$ from the wire. Since the wire is at a distance of $0.5\text{ m}$ to the left of the center $P$, the distance $d_R$ of point $R$ from the center $P$ is: $d_R = 2\text{ m} - 0.5\text{ m} = 1.5\text{ m}$ Since $d_R > R_{\text{shell}}$, point $R$ lies outside the spherical shell. The potential due to the shell at point $R$ is: $V_{\text{shell}}(R) = \frac{1}{4\pi\epsilon_0} \frac{Q}{d_R}$ Substituting the given values: $V_{\text{shell}}(R) = (9 \times 10^9) \times \frac{10 \times 10^{-9}}{1.5} = 60\text{ V}$

Thus, the potential difference due to the spherical shell is: $\Delta V_{\text{shell}} = V_{\text{shell}}(P) - V_{\text{shell}}(R) = 90\text{ V} - 60\text{ V} = 30\text{ V}$

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2. Contribution from the Infinitely Long Thin Wire

The electric field at a distance $r$ from an infinitely long wire with linear charge density $\lambda = 5\text{ nC/m} = 5 \times 10^{-9}\text{ C/m}$ is: $E(r) = \frac{\lambda}{2\pi\epsilon_0 r}$

The potential difference between points $P$ and $R$ due to the wire, where $r_P = 0.5\text{ m}$ and $r_R = 2\text{ m}$, is given by: $\Delta V_{\text{wire}} = V_{\text{wire}}(P) - V_{\text{wire}}(R) = \int_{r_P}^{r_R} E(r) \, dr = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{r_R}{r_P}\right)$

Using $\frac{1}{2\pi\epsilon_0} = 2 \times \frac{1}{4\pi\epsilon_0} = 2 \times (9 \times 10^9) = 18 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$: $\Delta V_{\text{wire}} = (18 \times 10^9) \times (5 \times 10^{-9}) \times \ln\left(\frac{2}{0.5}\right)$ $\Delta V_{\text{wire}} = 90 \times \ln(4) = 90 \times 2\ln(2)$

Given $\ln 2 = 0.7$: $\Delta V_{\text{wire}} = 180 \times 0.7 = 126\text{ V}$

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3. Total Potential Difference

By the principle of superposition, the total potential difference between points $P$ and $R$ is: $|V_P - V_R| = \Delta V_{\text{shell}} + \Delta V_{\text{wire}}$ $|V_P - V_R| = 30\text{ V} + 126\text{ V} = 156\text{ V}$

156