Question

A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an electric field.
A) What are the magnitude and direction of the electric field at the point in question? (Answer in N/C)

C) What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?(Answer in N)

Answer 1

Explanation:

Given that,

Charge acting on the object, $q=-4\ nC=-4\times 10^{-9}\ C$

Force acting on the object, $F=19\ nC=19\times 10^{-9}\ C$ (in downward direction)

(a) The electric force acting in the electric field is given by :

$F=qE$

E is the electric field

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

$E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}$

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, $q=1.6\times 10^{-19}\ C$

The force acting on the proton is :

$F=qE$

$F=1.6\times 10^{-19}\times 4.75$

$F=7.6\times 10^{-19}\ N$

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.