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Electric Fields & Force

  • #1
A small object carrying a charge of -5.50 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an electric field.

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

Magnitude & Direction of the Electric Field = 4.36 N/C upward





F = k (Q1Q2/r^2)
F = |q|E
k = 9E9





F = |q|E
= 5.5E-9 x 4.36
= 2.398E-8
 

Answers and Replies

  • #2
16
0
"Magnitude & Direction of the Electric Field = 4.36 N/C upward"


I think this is correct so far, and then all you need to do is multiply the E by the charge of a proton.

There is no need for k.

I think u pretty much got it, maybe # error.
 
  • #3
Shooting Star
Homework Helper
1,975
4
F = |q|E
= 5.5E-9 x 4.36
= 2.398E-8[/b]
If you have found the field yourself, then what do you think should put in place of q?
 

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