Question

An infinitely long thin wire, having a uniform charge density per unit length of \(5 \, \text{nC/m}\), is passing through a spherical shell of radius \(1 \, \text{m}\), as shown in the figure. A \(10 \, \text{nC}\) 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 \_\_\_\_. \[ \text{(Given: In SI units } \frac{1}{4\pi\epsilon_0} = 9 \times 10^9, \ln 2 = 0.7). \] Ignore the area pierced by the wire.
\includegraphics[width=0.3\linewidth]{ph12.png}

Hint:

When calculating potential differences in systems with multiple charge distributions, compute contributions from each separately and sum them up.

Answer 1

The potential difference consists of two components: 1. Due to the line charge, 2. Due to the spherical shell charge. 1. Potential difference due to the line charge: The electric field due to a line charge is: \[ E_{\text{line}} = \frac{\lambda}{2\pi\epsilon_0 r}. \] The potential difference is given by: \[ \Delta V_{\text{line}} = \int_{0.5}^{2} \frac{\lambda}{2\pi\epsilon_0 r} \, dr = \frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{2}{0.5}\right). \] Simplify: \[ \Delta V_{\text{line}} = \frac{\lambda}{2\pi\epsilon_0} \ln(4). \] 2. Potential difference due to the spherical shell charge: The potential at a distance \(r\) from a spherical shell with total charge \(Q\) is: \[ V = \frac{Q}{4\pi\epsilon_0 R} \quad \text{(outside the shell)}. \] Thus, the potential difference is: \[ \Delta V_{\text{sphere}} = \frac{Q}{4\pi\epsilon_0} \left(\frac{1}{R} - \frac{1}{2R}\right) = \frac{Q}{4\pi\epsilon_0} \cdot \frac{1}{2R}. \] Net potential difference: \[ \Delta V_{\text{net}} = \Delta V_{\text{line}} + \Delta V_{\text{sphere}}. \] Substitute values: \begin{aligned} \Delta V_{\text{line charge}} &= \int_{0.5}^{2} \frac{\lambda}{2\pi \epsilon_0 r} \, dr = \frac{\lambda}{2\pi \epsilon_0} \ln 4
\Delta V_{\text{sphere}} &= \frac{1}{4\pi \epsilon_0} \left( \frac{Q}{R} - \frac{Q}{2R} \right) = \frac{1}{4\pi \epsilon_0} \frac{Q}{2}
\Delta V_{\text{net}} &= \frac{\lambda}{2\pi \epsilon_0} \ln 4 + \frac{1}{4\pi \epsilon_0} \frac{Q}{2} = 171 \text{ volts} \end{aligned} \[ \Delta V_{\text{net}} = 171 \, \text{V}. \]