JEE Main 2020 · 8 January · morning · Single correct · +4 / −1
In finding the electric field using Gauss Law the formula | E⃗ | = q_(enc)/(ε ₀| A |) is applicable. In the formula ε ₀ is permittivity of free space, A is the area of Gaussian surface and q enc is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?
- (a) Only when | E⃗ | = constant on the surface.
- (b) For any choice of Gaussian surface.
- (c) Only when the Gaussian surface is an equipotential surface.
- (d) Only when the Gaussian surface is an equipotential surface and | E⃗ | is constant on the surface.
Answer and solution
Answer (d) Only when the Gaussian surface is an equipotential surface and | E⃗ | is…
Solution
Physics sir · 4 steps
To pull E out of the flux integral, E must be normal to the surface everywhere and the same size everywhere.
- Start from Gauss. Gauss's law in full is a surface integral. The dot product means only the part of E⃗ along the outward normal counts at each patch.
∮ E⃗ · dA⃗ = q_(enc)/ε₀
- Direction condition. E⃗ · dA⃗ becomes |E⃗| dA only if E⃗ is along the normal at every point. The field is always perpendicular to an equipotential surface, so the Gaussian surface must be equipotential.
- Magnitude condition. Even then, |E⃗| can come outside the integral only if it has the same value everywhere on the surface. Then ∮ |E⃗| dA = |E⃗| |A|.
- So. Both conditions are needed together for |E⃗| = q_(enc)/(ε₀ |A|), which is option (d).
The idea to keep. Gauss's law always holds, but it gives E directly only when the surface is chosen so that E is normal to it and constant in size on it. That is why we pick spheres, cylinders and pillboxes.
Answer from the NTA final answer key. Solution by PrepWiser.