# Coulomb law - 3 point charges

1. Apr 3, 2013

### Roodles01

1. The problem statement, all variables and given/known data
2 questions regarding the answer I have been given for this problem. Attachments are the problem & relevant worked answer I disagree with.

Problem
Three charges are arranged in the xy-plane as shown in attachment. A charge Q is at the point A with (x, y) coordinates (–2a, 4a), a charge –Q is at the point B which is the origin (0, 0), and a charge 2Q is at point C with coordinates (3a, 4a).

Find an expression for the force FC on the charge at C due to the other two charges and hence determine the magnitude |FC| of the force and the unit vector that specifies the direction of FC

2. Relevant equations
Coulomb's law
F = q1 q2 / 4 ∏ ε0 r123 rhat12

F = ke q1 q2 / r123 * rhat

where ke = 1 / 4 ∏ ε0

3. The attempt at a solution
FCA = 1/4∏ε0 qC qA / rCA3 * rCA

FCA = ke 2Q Q / (5a)3 * (5a ex)
OK, but then goes to

FCA = ke Q2 / a2 * 2/53 * (5a ex)

probably simple algebra but can't see how (5a)3 goes to a2 &53

Next query is regarding
FCB = ke * 2Q * -Q / (5a)3 * (3aex + 4aey)

How can the denominator be (5a)3 too, surely this is √20
Please assure me (or otherwise) that it's not (5a)3

Thank you

#### Attached Files:

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• ###### coulomb problem 02.JPG
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Last edited: Apr 3, 2013
2. Apr 3, 2013

### ap123

There's an "a" in the numerator which leaves a2 in the denominator.

Remember that rCA is the vector from C to A and is not the position vector of A

3. Apr 3, 2013

### Roodles01

Thank you.
Yes, easy points to remedy.