Question

A positive test charge of 5.00 E-5 C is placed in an electric field. The force on it is 0.751 N. The magnitude of the electric field at the location of the test charge is

Answer 1

We know that formula of force on charge is given by-

F= qE where q = charge , E=electric field

so $E=\dfrac{F}{q}=\dfrac{0.751}{5\times 10^{-5}}$ = $15020\ N\ C^{-1}$

electric field at the location E =$15020\ N \ C^{-1}$