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A semi-infinite line of charge (a rod extending from a point A to infinity) has uniform linear charge density lambda. Point P is located at perpendicular distance r from the end A of the rod (P lies level with the end, so the rod extends along AB starting at the near end). What is the electric field at P?
- 2*lambda² / ((4*pi*epsilon0*r)²), directed at 45 deg with AB
- (sqrt(2)*lambda) / (4*pi*epsilon0*r²), directed at 45 deg with AB
- (sqrt(2)*lambda) / (4*pi*epsilon0*r), directed at 45 deg with AB
- (sqrt(2)*lambda) / (4*pi*epsilon0*r), directed perpendicular to AB
Correct answer: (sqrt(2)*lambda) / (4*pi*epsilon0*r), directed at 45 deg with AB
Solution
For a semi-infinite line charge, at a point at perpendicular distance r from the end, the component of the field perpendicular to the rod is E_perp = lambda/(4*pi*epsilon0*r) and the component parallel to the rod is E_par = lambda/(4*pi*epsilon0*r). Since the two components are equal, the resultant makes 45 deg with the rod (AB) and has magnitude sqrt(E_perp² + E_par²) = sqrt(2)*lambda/(4*pi*epsilon0*r).
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