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JEE Advanced2017PhysicsElectrostaticsActual

A point charge + Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct?

Options

  1. AThe circumference of the flat surface is an equipotential
  2. BThe electric flux passing through the curved surface of the hemisphere is - Q 2 ε 0 1 - 1 2
  3. CTotal flux through the curved and the flat surfaces is Q ε 0
  4. DThe component of the electric field normal to the flat surface is constant over the surface

Correct answer

A. The circumference of the flat surface is an equipotential

Step-by-step solution

Every point on circumference of flat surface is at equal distance from point charge Hence circumference is equipotential. Flux passing through curved surface = - flux passing through flat surface. d ϕ t h r o u g h t h e r i n g = E cos ⁡ θ . d A = 1 4 π ∈ 0 Q r 2 + R 2 R R 2 + r 2 . 2 π r d r ∴ ∫ d ϕ = Q R 4 π ∈ 0 2 π ∫ 0 R r d r R 2 + r 2 3 2 = q 2 ∈ 0 1 - 1 2 ∴ Flux through curved surface = - q 2 ∈ 0 1 - 1 2 Note: Flux through surface can be calculated using concept of solid angle. Ω = 2 π 1 - cos ⁡ θ = 2 π 1 -

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