Calculating Net Force and Electric Field Using Coulomb's Law - Homework Question

[kurt1992]

Homework Statement

This is the charge distribution:B
|
|
| 2.0 m
|
|
A------------C
2.0 m

A = 2.0*10^-5 C
B = -3.0*10^-5 C
C = -3.0*10^-5 C

a) Find the net force on the charge

b) what is the net electric field acting on the charge

Homework Equations

FE=kq1q2/r^2

electric field = Fnet/q

The Attempt at a Solution

a) The charges acting on A are equal and at equal distance so they are then same.

F_E=(9.0*10^9)(2.0*10^-5)(-3.0*10^-5)/2^2

F_E=5.4/4

F_E=1.35 N

Each charge has a net force of 1.35 Newtons on charge 1. However, force is a vector quantity so the vectors have to have the same direction in order to add them. (The forces are the same so we can simply multiply the hypotenuse of the electric force by 2 to find the net force on particle 1.

(F_E/cos45)(2)=Fnet

3.81838=Fnet

Direction N45E

(is this correct? my books solution to the problem involves adding x and y components which make a very confusion solution.)

b)

ε=3.81838/(2.0*10^-5)
ε=190919 N/C

 

[haruspex]

(F_E/cos45)(2)=Fnet

That's not right. Is that what you meant to write?

 

[kurt1992]

yes, i thought adding the vectors together at 45 degrees would be the same as adding x and y components and then using trigonometry to find the magnitude.
Is it not equivalent?

 

[haruspex]

yes, i thought adding the vectors together at 45 degrees would be the same as adding x and y components and then using trigonometry to find the magnitude.
Is it not equivalent?

Yes, if done correctly. Your answer was almost right, so I wondered if it was just a typo. If it's what you meant to post, please post your full working.