A small part of `dl` length is removed from a ring having charge per unit length `lamda`. Find electric field at centre due to remaining ring.

Options

A. `E=(K(lamda)dl)/(R^(2))` (towards `-x`)
B. `E=(K(lamda)dl)/(R^(2))` (towards `+x`)
C. `E=(Klamdadl)/(2R^(2))` (towards `-x`)
D. `E=(Klamdadl)/(2R^(2))` (towards `+x`)

Answer

B

Explanation

`(k(Q))/(R^(2))=(-Klamdadl)/(R^(2))`