JEE Main 2025 Jan 29 Shift 2, Physics Q5: Gauss's Law

JEE Main2025Jan 29, Shift 2Physics

A point charge causes an electric flux of $- 2 \times 1 0^{4}$ $Nm^{2} C^{- 1}$ to pass through a spherical Gaussian surface of $8.0$ cm radius, centred on the charge. The value of the point charge is :

(Given $ε_{0} = 8.85 \times 1 0^{- 12} C^{2} N^{- 1} m^{- 2}$ )

A
$- 17.7 \times 1 0^{- 8}$ C
B
$- 15.7 \times 1 0^{- 8}$ C
C
$17.7 \times 1 0^{- 8}$ C
D
$15.7 \times 1 0^{- 8}$ C

Solution

By Gauss's law, $ϕ = \frac{q}{ε _{0}}$ , so $q = ε_{0} ϕ$ . The radius is irrelevant when the charge sits at the centre of the Gaussian surface, because the entire flux is enclosed regardless.

$q = (8.85 \times 1 0^{- 12}) (- 2 \times 1 0^{4}) = - 17.7 \times 1 0^{- 8}$ C.

The negative sign is consistent with the negative flux given.

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JEE Main 2025 Jan 29 Shift 2, Physics Q5: Gauss's Law

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