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JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2024 · 29 January · Morning Shift · Physics · Electrostatics

Two charges of 5Q and - 2Q are situated at the points ( 3a, 0 ) and...

Two charges of 5Q5Q and −2Q- 2Q are situated at the points (3a,0)( 3a, 0 ) and (−5a,0)( - 5a, 0 ) respectively. The electric flux through a sphere of radius 4a4a having centre at origin is:Two charges of 5Q and - 2Q are situated at the points ( 3a, 0 ) and ( - 5a, 0 ) respectively. The electric flux through

Options

  1. A2Qε0\dfrac{2Q}{{ε}_{0}}
  2. B5Qε0\dfrac{5Q}{{ε}_{0}}
  3. C7Qε0\dfrac{7Q}{{ε}_{0}}
  4. D3Qε0\dfrac{3Q}{{ε}_{0}}

Correct answer

B. 5Qε0\dfrac{5Q}{{ε}_{0}}

Step-by-step solution

Two charges of 5Q and - 2Q are situated at the points ( 3a, 0 ) and ( - 5a, 0 ) respectively. The electric flux through

As can be seen from the figure above, only the 5Q5Q charge is inside the spherical region.

The formula to calculate the flux through the surface is given by

ϕ=qencε0...(1)ϕ = \dfrac{{q}_{enc}}{{ε}_{0}} . . . \left(1\right)

From equation (1)( 1 ), the flux through the given surface can be calculated as follows:

ϕ=5Qε0ϕ = \dfrac{5Q}{{ε}_{0}}

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