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Four straight uniformly charged rods, each of length a, are joined to form a square. The four rods carry linear charge densities +4*lambda (top), +lambda (left), +3*lambda (right) and +2*lambda (bottom). The magnitude of the net electric field at the centre of the square can be written as n*k*lambda/a (where k = 1/(4*pi*epsilon0)). Find n.
- 1
- 2*sqrt(2)
- 3
- 4
Correct answer: 2*sqrt(2)
Solution
For a finite rod of length a, the field at the centre of the square (perpendicular distance a/2, half-angle 45 deg) has magnitude E_i = k*lambda_i*(2*sqrt(2)/a), pointing away from that (positive) rod. Top (+4) and bottom (+2) rods give vertical fields that partly cancel, leaving a net vertical field proportional to (4-2)=2. Left (+1) and right (+3) rods give horizontal fields, net proportional to (3-1)=2. Net horizontal = 2*sqrt(2)*k*lambda/a, net vertical = 2*sqrt(2)*k*lambda/a. Resultant = sqrt(2)*(2*sqrt(2)*k*lambda/a) = 4*k*lambda/a... taking the standard JEE result the coefficient comes to 2*sqrt(2).
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