Two point charges. Given one find the other?

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Homework Help Overview

The problem involves two point charges located 35 cm apart in a vacuum, with one charge known to be a positive 16 μC at point A. The field strength at point X, situated 12 cm from A, is given as zero, leading to the need to determine the charge at point B.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the charges and the distances involved, using Coulomb's law to set up equations. Some express uncertainty about how to proceed without knowing the distance to charge B, while others attempt to manipulate the equations to isolate the unknown charge.

Discussion Status

Several participants have engaged in calculations and provided different interpretations of the equations. There is ongoing exploration of how to correctly apply the formulas to find the charge at B, with some guidance offered regarding the use of electric field strength and the correct substitution of known values.

Contextual Notes

Participants note the challenge of working with incomplete information and the implications of the field strength being zero at point X. There is also mention of a diagram that clarifies the positions of the charges, which may influence the calculations.

afrocod
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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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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
 
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.
 
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)
 
(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:
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).
 
Oh, I see. I really should have known that... I think I need some sleep.

Thank you for your patience and very clear explanation.
 

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