What will be the magnitude of electric field at point O as shown in the figure? Each side of the figure is l and perpendicular to each other?

Solution

Physics sir · 4 steps

Pair up opposite charges: the near four push along one diagonal and the corner charges push back.

1. Set up. Put O at the origin with x to the right and y up, and let E₀ = 1/4πε₀q/l². B, G, C and F are at distance l. A, D, H and E sit at the corners, at distance √2 l, so each gives E₀/2 per q.
2. Nearest four. Along x, B (+q) gives E₀ to the right and G (2q) gives 2E₀ to the left, so the net is E₀ left. Along y, C (2q) gives 2E₀ down and F (q) gives E₀ up, so the net is E₀ down. Together that is √2 E₀ pointing towards E, the lower-left.
3. Corner charges. A and H are both -q and sit opposite each other, so their fields cancel. On the other diagonal, E (2q) pushes towards the upper-right with E₀ and D (+q) pushes towards the lower-left with E₀/2. The net is E₀/2 towards the upper-right.

   E_O = √2 E₀ - E₀/2 = 1/4πε₀q/2l²(2√2-1)

4. So. Both results lie along the same diagonal in opposite directions, so they subtract. The magnitude at O is 1/4πε₀q/2l²(2√2-1), which is option (b).

The idea to keep. With many point charges, first cancel equal charges placed opposite each other, then pair up the rest along each line. Add the leftover vectors only at the end.

Answer from the NTA final answer key. Solution by PrepWiser.