JEE Main 2023 · 30 January · evening · Single correct · +4 / −1
As shown in the figure, a point charge Q is placed at the centre of conducting spherical shell of inner radius a and outer radius b. The electric field due to charge Q in three different regions I, II and III is given by:
(I: r < a, II: a < r < b, III: r>b )
- (a) E_I=0, E_(I I)=0, E_(I I I) ≠ 0
- (b) E_I ≠ 0, E_(I I)=0, E_(III)=0
- (c) E_I ≠ 0, E_(I I)=0, E_(III) ≠ 0
- (d) E_I=0, E_(I I)=0, E_(I I I)=0
Answer and solution
Answer (c) E_I ≠ 0, E_(I I)=0, E_(III) ≠ 0
Solution
Physics sir · 2 steps
Draw a Gaussian sphere in each region and ask how much charge it encloses; the conductor sorts out the rest.
- Induced charges. The field inside the metal of a conductor is zero, so a charge -Q collects on the inner surface (radius a) to cancel Q. The shell is neutral, so +Q appears on the outer surface (radius b). The question means the net field in each region.
∮ E⃗· dA⃗ = q_(enc)/ε₀ ⇒ E = q_(enc)/(4πε₀ r²)
Region Charge inside a Gaussian sphere of radius r Field I: r < a Q Q/(4πε₀ r²) ≠ 0 II: a < r < b Q + (-Q) = 0 0 (inside the metal) III: r > b Q - Q + Q = Q Q/(4πε₀ r²) ≠ 0 - So. The field is non-zero in region I, zero in region II and non-zero again in region III. That is option (c).
The idea to keep. A conductor kills the field only inside its own metal. It does not hide the charge from the outside world: a neutral shell passes the enclosed charge on to its outer surface.
Answer from the NTA final answer key. Solution by PrepWiser.