A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length . A small length dL of the wire is cut off. Find the electric field at the centre due to remaining wire.

Options

A. `(kQdL)/(4pia^3)`
B. `(kQdL)/(2pia^3)`
C. `(kQdL)/(2Lpia^3)`
D. `(kQdL)/(4pia^2)`

Answer

B

Explanation

` E=1/(4piepsilon_0) (dq)/a^2 `
` dq=Q/(2pia).dL `
` Therefore, E=1/(4piepsilon_0a^2) Q/(2pia)dL `
` (QdL)/(8pi^2epsilon_0a^3)`