
An infinitely long thin wire, having a uniform charge density per unit length of $5 \mathrm{nC} /…

Solution
due to wire
$\begin{aligned} & \mathrm{dV}=-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dx}} \\ & \int_{\mathrm{v}_{\mathrm{P}}}^{\mathrm{v}_{\mathrm{R}}} \mathrm{dV}=-\int_{0.5}^2 \frac{2 \mathrm{k} \lambda}{\mathrm{x}} \mathrm{dx} \\ & \mathrm{v}_{\mathrm{R}}-\mathrm{v}_{\mathrm{P}}=-2 \mathrm{k} \lambda \ln \frac{2}{0.5} \\ & =-2 \times 9 \times 10^9 \times 3 \times 10^{-9} \times 2 \times 0.7=-126 \mathrm{~V}\end{aligned}$
due to sphere
$\begin{aligned} & \mathrm{v}_{\mathrm{R}}-\mathrm{v}_P=\frac{\mathrm{kQ}}{2}-\frac{\mathrm{kQ}}{1}=-\frac{\mathrm{kQ}}{2}=\frac{-9 \times 10^9 \times 10 \times 10^{-9}}{2} \\ &=-45 \mathrm{~V} \\ & \mathrm{v}_{\mathrm{R}}-\mathrm{v}_{\mathrm{P}}=-126-45=-171 \mathrm{~V} \\ & \mathrm{v}_{\mathrm{P}}-\mathrm{v}_{\mathrm{R}}=171 \mathrm{~V}\end{aligned}$
^Asked in: JEE Advanced 2024 (Paper 2)