Question

The electric force experienced by a charge of 1*10^-6c is 1.5*10^-3n.Find the magnitude of the electric field at the position of the charge

Answer 1

Given,

Force, F=1.5*10^-3 N

Charge, q=1*10^-6 C

Force = charge x electric field

F=qE

E=F/q

E=(1.5*10^-3)/(1*10^-6)

E=1500 N/C

hence, the electric field at position of charge is 1500 N/C

Answer 2

The magnitude of the electric field at the position of the charge is 1500 N/C.

Explanation:

It is given that,

Charge, $q=10^{-6}\ C$

Electric force, $F=1.5\times 10^{-3}\ N$

To find,

The magnitude of the electric field at the position of the charge.

Solution,

The electric force per unit of electric charge is called electric field at the position of the charge. It is given by :

$E=\dfrac{F}{q}$

$E=\dfrac{1.5\times 10^{-3}\ N}{10^{-6}\ C}$

E = 1500 N/C

So, the magnitude of the electric field at the position of the charge is 1500 N/C.

Learn more

Electric field

https://brainly.in/question/14398255