# Force on point charge A?

## Homework Statement

What is the force on the 1.0nC charge in the middle of the figure due to the four other charges?

http://filesmelt.com/dl/question.JPG [Broken]

## Homework Equations

F = K*q1*q2 / r^2
K = 8.99x10^9

## The Attempt at a Solution

Initial observation: The symmetry of the charges means that Fy = 0 so I only need to worry about Fx.

F of P1 on A = - K*q[P1]*q[A] / r^2

r^2 = a^2 + b^2 = 0.5^2 + 0.5^2 = 0.25 + 0.25 = 0.5

K*q[P1]*q[A] / 0.5 = - K*q[P1]*q[A]*2

Second observation: The force of P1 on A is equal to the force of P2 on A

Therefore the force of both P1 and P2 on A is

- K*q[P1]*q[A]*2*2

Third observation: The force of N1 and N2 on A is equal to the force of P1 and P2 on A

Therefore the force of N1, N2, P1 and P2 on A (which is Fx) is

- K*q[P1]*q[A]*2*2*2 = - K*q[P1]*q[A]*8

- (8.99x10^9)(1x10^-19)(2x10^-19)*8 = 0.0000000000000000000000000014384

Where did I go wrong? Thanks in advance!

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## Answers and Replies

supratim1
Gold Member
you have not taken the x-component of the forces. multiply by sin45.