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Electrostatics question

2025 · 29 Jan · Shift 2 · Q67
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Electrostatics question

2025 · 29 Jan · Shift 2 · Q67

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A point charge causes an electric flux of −2×104Nm2C−1-2 \times 10^4 \mathrm{Nm}^2 \mathrm{C}^{-1}−2×104Nm2C−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}ϵ0​=8.85×10−12C2 N−1 m−2 )
  1. A
    17.7×10−8C17.7 \times 10^{-8} \mathrm{C}17.7×10−8C
  2. B
    −17.7×10−8C-17.7 \times 10^{-8} \mathrm{C}−17.7×10−8C
  3. C
    15.7×10−8C15.7 \times 10^{-8} \mathrm{C}15.7×10−8C
  4. D
    −15.7×10−8C-15.7 \times 10^{-8} \mathrm{C}−15.7×10−8C
View written solutionFree

Correct answer: B

  1. Use Gauss's law

For a closed surface,

Φ=qencε0\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}Φ=ε0​qenc​​

So,

qenc=ε0Φq_{\text{enc}} = \varepsilon_0 \Phiqenc​=ε0​Φ
  1. Substitute the given values

Given:

Φ=−2×104 N m2C−1,ε0=8.85×10−12 C2N−1m−2\Phi = -2 \times 10^4\ \text{N m}^2\text{C}^{-1}, \qquad \varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}Φ=−2×104 N m2C−1,ε0​=8.85×10−12 C2N−1m−2

Therefore,

q=(8.85×10−12)(−2×104)q = (8.85 \times 10^{-12})(-2 \times 10^4)q=(8.85×10−12)(−2×104)
  1. Calculate
q=−17.7×10−8 Cq = -17.7 \times 10^{-8}\ \text{C}q=−17.7×10−8 C

Equivalently,

q=−1.77×10−7 Cq = -1.77 \times 10^{-7}\ \text{C}q=−1.77×10−7 C
  1. Identify the correct option

The value matches:

B: −17.7×10−8 C\boxed{\text{B: } -17.7 \times 10^{-8}\ \text{C}}B: −17.7×10−8 C​
  1. Why radius is irrelevant

The radius of the spherical Gaussian surface does not affect the total flux through it. By Gauss's law, total flux depends only on the enclosed charge.

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