Asmall object carrying a charge of -2.50 nc is acted upon by a downward force of 18.0 nn when placed at a certain point in an electric field

B.What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"

Solution:

A) A charge q under an electric field of intensity E will experience a force F  equal to:

In our problem we have  and , so we can find the magnitude of the electric field:

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.

B) The proton charge is equal to

Therefore, the magnitude of the force acting on the proton will be

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.

a=d/t

a="triangle"v/t

explanation:

a = acceleration