# Homework Help: Two point charges. Given one find the other?

1. Mar 21, 2012

### afrocod

1. The problem statement, all variables and given/known data

Two point charges are situated at A and B a distance of 35cm apart in a vacuum. The charge A is a positive charge of 16 μC and the field strength at X, a distance of 12 cm from A, is zero. Calculate:

(i) The field strength at X due to the charge at A. (I found this to be 9.9 x 10^6 NC^-1. Just in case this is relevant to the next question.)

(ii) The charge at B.

2. Relevant equations

Permittivity of free space ε0 = 8.9 x 10^-12 F m^-1

F = Q.q / 4πεd^2

3. The attempt at a solution
It seems like I don't have enough information but here's an attempt anyway

F = (16 x 10^-6) / 4π(8.9 x 10^-12)(0.35)^2 = 1.2 x 10^6

The answer is 58.8 NC so im way off.

2. Mar 21, 2012

### pcm

field at X is zero.
so,q(A)/(AX)^2 =q(B)/(XB)^2
you know q(A),AX,XB.
solve it to get q(B)=58.8*10^-6

3. Mar 21, 2012

### afrocod

How do I find XB if I dont know B?

F = q(B) / 4πε0(0.23)^2 = q(B) / 5.91637

With this I get a cancellation of q(B) and can't find it.

Thanks for the formula, I have never seen it before.

4. Mar 21, 2012

### emailanmol

A and B are lying on a line at a distance of 35 cms.

Think of A lying on 0 on the number line and B at 35.

Now X can be at 12 or -12.

Field strength at a point is given by kq/r^2( You have written the equation of force.) You can guess the direction depending on sign of charge.

At X field is 0.

So Q(A)/144 = Q(B)/R^2

where R can have two values.

You know Q(A).
Plug in and find two values of Q(B)

5. Mar 21, 2012

### afrocod

(9.9 x 10^6) / 144 = Q(B) / 529

Q(B) = 36.37 x 10^6

It's definitely +12, there's a diagram that shows X between the two points. Sorry I forgot to mention.

I thought your explanation was crystal clear so I'm really at a loss as to why this is yielding the wrong answer.

Last edited: Mar 21, 2012
6. Mar 21, 2012

### emailanmol

Hey.

(9.9 x 10^6) / 144 = Q(B) / 529

This step is wrong.

Instead of the 9.9x10^6 you have to actually plug in Q (A) which is charge of A.

See Coulombs law tells us Force is kq(1)q(2)/r^2 ( with appropriate direction).

Now electric field strength of a charged particle having charge Q at a distance r is

kQ/r^2

Its like plugging in q(2) as 1 in force equation if we want to find electric field due to q(1).

7. Mar 21, 2012

### afrocod

Oh, I see. I really should have known that... I think I need some sleep.

Thank you for your patience and very clear explanation.