Force on Charge Q at Corner of Square

[Bryon]

Homework Statement

Four charges q1 = q3 = -q and q2 = q4 = +q, where q = 4 µC, are fixed at the corners of a square with sides a = 1.3 m.

q1 M' q4







q2 q3




(b) Find the force on a test charge Q = -0.3 µC placed at the midpoint M' of the top side of the square

Homework Equations

Coulombs law: F = (k*q1*q2)/r^2

The Attempt at a Solution

Fq1M' = (k(4uC)(0.3uC))/(0.65^2) = 0.025533728
Fq4M' = 0.025533728

theta = tan^(-1)(0.65/1.3) = 26.56505118 degrees
90-26.56505118 = 63.43494882 degrees

Fq2M'x = -0.025533728(cos(63.43494882)) = -0.01141903
Fq3M'x = -0.01141903

FM',net = 2(-0.01141903) + 2(0.025533728) = 0.028229396N

Which is wrong of course. Where did I mess this up at?

 

[Delphi51]

Fq2M'x = -0.025533728(cos(63.43494882))

You can't use the .0255 again because q2 and Q are not .65 m apart.

 

[Bryon]

oh yeah...that was a really silly mistake on my part. Thanks!

 

[Delphi51]

Most welcome, Bryon.