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Physics · Electrostatics

JEE Main 2025 — 29 January, Evening Shift — Question 21

A point charge causes an electric flux of −2×104-2 \times 10^{4} Nm2C−1\mathrm{Nm}^{2} \mathrm{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 ϵ0=8.85×10−12C2 N−1 m−2\epsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} )

  1. Option A:

    −17.7×10−8C-17.7 \times 10^{-8} \mathrm{C}

    Correct
  2. Option B:

    −15.7×10−8C-15.7 \times 10^{-8} \mathrm{C}

  3. Option C:

    17.7×10−8C17.7 \times 10^{-8} \mathrm{C}

  4. Option D:

    15.7×10−8C15.7 \times 10^{-8} \mathrm{C}

Answer: A

Step-by-step solution

ϕ=−2×104Nm2C\phi=-2 \times 10^{4} \frac{\mathrm{Nm}^{2}}{\mathrm{C}}

r=8.0 cm\mathrm{r}=8.0 \mathrm{~cm}

ϕ=qϵ0⇒q=ϵ0ϕ\phi=\frac{\mathrm{q}}{\epsilon_{0}} \Rightarrow \mathrm{q}=\epsilon_{0} \phi

=(8.85×10−12)×(−2×104)=\left(8.85 \times 10^{-12}\right) \times\left(-2 \times 10^{4}\right)

q=−17.7×10−8Cq=-17.7 \times 10^{-8} \mathrm{C}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
29 January, Evening Shift
Subject
Physics
Chapter
Electrostatics
Topic
Electric flux and Gauss's Law

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