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Electrostatics question

2022 · 27 Jun · Shift 2 · Q55
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Electrostatics question

2022 · 27 Jun · Shift 2 · Q55

JEE MainPhysicsElectrostaticsMCQ+4 / −1
If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be : JEE Main 2022 (Online) 27th June Evening Shift Physics - Electrostatics Question 124 English
  1. A
    q∈0{q \over {{ \in _0}}}∈0​q​
  2. B
    q2∈0{q \over {{ 2\in _0}}}2∈0​q​
  3. C
    q4∈0{q \over {{ 4\in _0}}}4∈0​q​
  4. D
    q2π∈0{q \over {{ 2\pi\in _0}}}2π∈0​q​
View written solutionFree

Correct answer: B

  1. Use Gauss’s law for a closed surface

If a point charge qqq is placed at the centre of a closed hemispherical surface (curved hemisphere + flat circular face), then the total electric flux through the entire closed surface is

Φtotal=qε0\Phi_{\text{total}} = \frac{q}{\varepsilon_0}Φtotal​=ε0​q​
  1. Symmetry of the arrangement

The charge is at the centre of the sphere of which the hemisphere is a half. The closed hemispherical surface consists of:

  • one curved hemispherical part
  • one flat circular part

Now imagine a full sphere centered at the charge. The total flux through the full sphere is

Φsphere=qε0\Phi_{\text{sphere}} = \frac{q}{\varepsilon_0}Φsphere​=ε0​q​

By symmetry, flux through the curved hemispherical surface is half of this:

Φcurved=12⋅qε0=q2ε0\Phi_{\text{curved}} = \frac{1}{2}\cdot \frac{q}{\varepsilon_0} = \frac{q}{2\varepsilon_0}Φcurved​=21​⋅ε0​q​=2ε0​q​
  1. Flux through the flat surface

For the closed hemispherical surface,

Φtotal=Φcurved+Φflat\Phi_{\text{total}} = \Phi_{\text{curved}} + \Phi_{\text{flat}}Φtotal​=Φcurved​+Φflat​

So,

qε0=q2ε0+Φflat\frac{q}{\varepsilon_0} = \frac{q}{2\varepsilon_0} + \Phi_{\text{flat}}ε0​q​=2ε0​q​+Φflat​

Hence,

Φflat=q2ε0\Phi_{\text{flat}} = \frac{q}{2\varepsilon_0}Φflat​=2ε0​q​
  1. Direction/sign note

If outward normal of the flat face is taken, the flux through the flat surface is actually negative because electric field enters the closed surface through that face. But since the options give only magnitude, the required answer is

q2ε0\boxed{\frac{q}{2\varepsilon_0}}2ε0​q​​
  1. Option check
  • A: qε0\dfrac{q}{\varepsilon_0}ε0​q​ ❌
  • B: q2ε0\dfrac{q}{2\varepsilon_0}2ε0​q​ ✅
  • C: q4ε0\dfrac{q}{4\varepsilon_0}4ε0​q​ ❌
  • D: q2πε0\dfrac{q}{2\pi\varepsilon_0}2πε0​q​ ❌
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