JEE Advanced 2024 Paper 2, Physics Section 3 Q5: Electric Potential

JEE Advanced2024Paper 2Physics Section 3

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 $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9}$ , $ln 2 = 0.7$ . Ignore the area pierced by the wire.]

Your answer

Solution

Potential difference contributions add from the two sources.

Line charge contribution between $r_{P} = 0.5$ m and $r_{R} = 2$ m:

$Δ V_{line} = \int_{0.5}^{2} \frac{λ}{2 π ε _{0} r} d r = \frac{λ}{2 π ε _{0}} ln 4 = 2 \cdot 9 \times 1 0^{9} \cdot 5 \times 1 0^{- 9} \cdot 2 ln 2.$

Using $ln 2 = 0.7$ : $Δ V_{line} = 2 \cdot 9 \cdot 5 \cdot 2 \cdot 0.7 = 126 V .$

Spherical shell contribution. Point $P$ is inside the shell (potential $= k Q / R = 9 \times 1 0^{9} \cdot 10 \times 1 0^{- 9} /1 = 90$ V), and point $R$ is outside at distance $2$ m (potential $= k Q / r = 90/2 = 45$ V).

$Δ V_{shell} = 90 - 45 = 45 V .$

Total: $∣ V_{P} - V_{R} ∣ = 126 + 45 = 171 V .$

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JEE Advanced 2024 Paper 2, Physics Section 3 Q5: Electric Potential

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