Two point charges. Given one find the other?

In summary, the charge at A has a negligible effect on the electric field at X, while the charge at B has a significant effect.
  • #1
afrocod
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0

Homework Statement



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.


Homework Equations



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

F = Q.q / 4πεd^2


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 I am way off.
 
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  • #2
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
How do I find XB if I don't 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
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
(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:
  • #6
afrocod said:
(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.


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

Thank you for your patience and very clear explanation.
 

FAQ: Two point charges. Given one find the other?

1. How do I find the magnitude of the other charge?

To find the magnitude of the other charge, you can use the formula Q = k * (q1 * q2) / d^2, where Q is the magnitude of the other charge, k is the Coulomb's constant, q1 and q2 are the magnitudes of the two known charges, and d is the distance between them.

2. What is the equation for finding the direction of the other charge?

The direction of the other charge can be found using the formula θ = tan-1 (y/x), where θ is the direction of the other charge, y is the distance between the two charges on the y-axis, and x is the distance between the two charges on the x-axis.

3. How can I determine the net force between two point charges?

The net force between two point charges can be determined using the formula F = k * (q1 * q2) / d^2, where F is the net force, k is the Coulomb's constant, q1 and q2 are the magnitudes of the two charges, and d is the distance between them.

4. Can the distance between the two charges affect the magnitude of the other charge?

Yes, the distance between the two charges can affect the magnitude of the other charge. As the distance increases, the magnitude of the other charge decreases, and vice versa.

5. How do I apply the principle of superposition to two point charges?

To apply the principle of superposition to two point charges, you can simply add the forces exerted by each individual charge on a third test charge. This will give you the net force on the test charge.

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