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 magnitu...

The correct numerical answer is 171. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Physics Question 12 Solution

Let $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ Nm $^2$ /C $^2$ . Wire charge density $\lambda = 5 \times 10^{-9}$ C/m. Shell charge $Q = 10$ \times $10^{-9}$ C.
The potential due to a thin shell is $V_s = \frac{kQ}{R}$ for $r \leq R$ and $V_s = \frac{kQ}{r}$ for $r \geq R$ .
The potential due to an infinite wire at distance $r$ is $V_w = -2k\lambda \ln(r) + C$ .
Point P is at $r_P = 0.5$ m (inside shell, $R=1$ m). Point R is at $r_R = 2$ m (outside shell).
$V_P = V_s(0.5) + V_w(0.5) = \frac{kQ}{1} - 2k\lambda \ln(0.5)$ .
$V_R = V_s(2) + V_w(2) = \frac{kQ}{2} - 2k\lambda \ln(2)$ .
Potential difference $\Delta V = |V_P - V_R| = |(\frac{kQ}{1} - \frac{kQ}{2}) + (2k\lambda \ln 2 - 2k\lambda \ln 0.5)|$ .
$\Delta V = |\frac{kQ}{2} + 2k\lambda \ln(\frac{2}{0.5})| = |\frac{kQ}{2} + 2k\lambda \ln 4| = |\frac{kQ}{2} + 4k\lambda \ln 2|$ .
Substitute values: $kQ = (9 \times 10^9)(10 \times 10^{-9}) = 90$ , $k\lambda = (9 \times 10^9)(5 \times 10^{-9}) = 45$ .
$\Delta V = |\frac{90}{2} + 4(45)(0.7)| = |45 + 180 \times 0.7| = |45 + 126| = 171$ Volts.