JEE Main 2021 · 24 February · morning · Single correct · +4 / −1
A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is -Q. The electric field at the centre of cube is :
- (a) 2Q/(3√3 π ε ₀a²)( x + y + z )
- (b) (- Q)/(3√3 π ε ₀a²)( x + y + z )
- (c) Q/(3√3 π ε ₀a²)( x + y + z )
- (d) (- 2Q)/(3√3 π ε ₀a²)( x + y + z )
Answer and solution
Answer (d) (- 2Q)/(3√3 π ε ₀a²)( x + y + z )
Solution
Physics sir · 4 steps
Write the -Q as +Q plus -2Q: the eight +Q charges cancel and only -2Q at the origin remains.
- Split the charge. Think of the charge at the origin as +Q and -2Q sitting together, since +Q + (-2Q) = -Q. Now you have +Q at all eight corners, plus an extra -2Q at the origin.
- Symmetry cancels. Eight equal charges at the corners give zero field at the centre. Each corner is paired with the opposite corner, the same distance away on the other side. So the whole field comes from the -2Q alone.
- Distance and direction. The centre is at (a/2, a/2, a/2), so r² = 3a²/4. The field of a negative charge points towards it, so here it points from the centre back to the origin, along -(x+y+z)/√3.
E⃗ = 1/4πε₀·2Q/(3a²/4)(-(x+y+z)/√3) = -2Q/(3√3 πε₀ a²)(x+y+z)
- So. The field at the centre has magnitude 2Q/(3πε₀ a²) and points towards the origin, which is option (d).
The idea to keep. When one charge in a symmetric arrangement is different, split it as (the usual charge) plus (the difference). The symmetric part cancels, so only the difference charge produces the field.
Answer from the NTA final answer key. Solution by PrepWiser.