Calculating Net Force: Equilateral Triangle with Point Charges

[psruler]

Hi,
Can anybody help me get started on this problem?

Point P2 is located 3.0 m away from each -Q (-Q = -40 nC) , forming an equilateral triangle with them.

Determine the net force (magnitude and direction) that would act on a small charge of +1.0 nC placed at P2.

Any hints?

Thanks!

 

[Beer-monster]

Draw a picture and use Coulomb's law. Make use of verical and horizontal components.

 

[Sirus]

Right. Draw the vector for the force that each other charge applies on the +1.0 nC charge (using Coulomb's law), then add them together in a vector addition to find the resultant.

 

[psruler]

The answer i got is: 4.4 x 10^-8N.
Can you verify if that's right?

Here are the steps I did to calculate that:

F12 = k(q1)(q2)/r^2 = [(8.99 x 10^9)(1.0 x 10^-9)(-40 x 10^-9)]/(3.0)^2

F12x = F12cos56 = (4.0 x 10^-8)(0.56) = 2.2 x 10^-8
F12y = F12sin56 = (4.0 x 10^-8)(0.83) = 3.3x 10^-8
F13 = 4.0 x 10^-8
F13x = 2.2 x 10^-8
F13y = -3.3 x 10^-8

F1x = F12x + F13x = (2.2 x 10^-8) + (2.2 x 10^-8) = 4.4 x 10^-8
F1y = F12y - F13y = 0
F= (F1x^2 + F1y^2)^1/2 = [(4.4 x 10^8)^2 + (0)^2]^1/2 = 4.4 x 10 ^-8N

THANKS!