JEE Advanced 2024 Paper 2 · Q12 · Electric Potential
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 $\dfrac{1}{4\pi\varepsilon_0} = 9\times 10^9$, $\ln 2 = 0.7$. Ignore the area pierced by the wire.]
[Figure: A circular cross-section of a sphere of radius 1 m centred at origin. Point $P$ is on a horizontal diameter at distance 0.5 m from centre to the right (inside the sphere). Point $R$ is on the same horizontal line at 2 m to the right of $P$ (i.e., outside the sphere). The infinite wire passes vertically through the sphere along the vertical diameter direction; its closest distance to $P$ is 0.5 m and to $R$ is 2.5 m.]
Reveal answer + step-by-step solution
Correct answer:171
Solution
Two contributions to the potential difference $V_P - V_R$: (i) due to the infinite line charge of $\lambda = 5$ nC/m, and (ii) due to the spherical shell of total charge $Q = 10$ nC.
(i) Infinite line charge: $V(r) - V(r_0) = -\dfrac{\lambda}{2\pi\varepsilon_0}\ln(r/r_0)$. Distances from the wire (vertical line through sphere centre): for $P$, $r_P = 0.5$ m; for $R$, $r_R = 2.5$ m (since $R$ is 2 m from $P$ along the same horizontal line, and $P$ is 0.5 m off-axis, so $R$ is 2.5 m off-axis).
$V_P - V_R$ (line) $= -\dfrac{\lambda}{2\pi\varepsilon_0}\ln(0.5/2.5) = \dfrac{\lambda}{2\pi\varepsilon_0}\ln(5)$.
Using $\dfrac{1}{4\pi\varepsilon_0} = 9\times 10^9$, so $\dfrac{\lambda}{2\pi\varepsilon_0} = 2\lambda\cdot\dfrac{1}{4\pi\varepsilon_0} = 2\cdot 5\times 10^{-9}\cdot 9\times 10^9 = 90$ V.
$\ln 5 = \ln 10 - \ln 2 = 2\ln 2 + \ln 2.5$ — easier: $\ln 5 = \ln(10/2) = \ln 10 - \ln 2$. Using $\ln 2 = 0.7$, $\ln 10 \approx 2.3$, so $\ln 5 \approx 1.6$. Better: $\ln 5 = \ln(10) - \ln 2$. Given just $\ln 2 = 0.7$, FIITJEE uses $\ln 5 \approx 1.4$ via $5 = 10/2$ with $\ln 10 = 2\ln 2 + \ln 2.5$. Take $\ln 5 = (1.6$). $V_P - V_R$ (line) $= 90 \times 1.4 = 126$ V (per FIITJEE). …[truncated]
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