Solving Magnetic Fields: Find Electric Field Zero Point

AI Thread Summary
The discussion revolves around calculating the point where the electric field is zero between two charges, 6e-5 C and -2e-5 C, placed 0.36 m apart. The user is attempting to set up the equations but struggles with the placement of the zero electric field point and the distances involved. It is clarified that the distance from the positive charge to the zero point should be expressed as (0.36 + r) instead of (0.36 - r). The user is advised to equate the electric fields from both charges at point P and correct the sign in their calculations to arrive at the correct answer of 0.49 m. The conversation highlights the importance of accurately defining distances in electric field problems.
Matt1234
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Homework Statement



Two small charges 6e-5 and -2e-5 C. are placed 0.36m apart. Calculate the following:
e) the point at which the magnitude of the electric field is zero.



Homework Equations



EF (field) = (k q) / r^2

where EF = 0

k = 9e9

The Attempt at a Solution



I know I am looking for r, but the way i setup the equation gives me 2 unknowns.
 
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Distance between the two charges is 0.36cm, so put the point where E=0 as a distance x from one charge, so the distance from the other charge would be?
When you get that, you will have the equation with one unknown.
 
im sorry i don't understand. if i set one r to 0.36 and solve for the other?

i tried that and i get 0.2 which is wrong. I am not 100% sure about this part "put the point where E=0 as a distance x from one charge"
 
Matt1234 said:
i tried that and i get 0.2 which is wrong. I am not 100% sure about this part "put the point where E=0 as a distance x from one charge"


+ _______________ -
(0.36cm)


+ ______ (E=0) _______ -
(x) (p) ??


if the distance from + to - is 0.36cm (top diagram), then what the '??' equal to in the second diagram? (something-x, what is that something?)

When you get that, can you find the electric field at point P (where E=0) due to the +ve charge in terms of x? Similarly, do the same for the electric field at point P due tot he -ve charge. Those two are equal, you can now find x.

EDIT: my diagrams are not coming out correctly. So for the second one, under the first line is 'x', under the E=0 is 'P' and under the second line is '??'
 
(0.36 - r) i believe is what I am looking for?

Making this question quadratic?

i got the root to be 0.56 which.

The answer is 0.49m

under the first i used:

(0.36^2) under the second i used ( x -0.36)^2 which made it quadratic.
 
Last edited:
Matt1234 said:
(0.36 - r) i believe is what I am looking for?

Making this question quadratic?

Yes.
 
ok I am quite confused as i keep getting 0.56 yet the answer is 0.49 m

ill show the teacher my work tom. thank you.
 
Last edited:
Matt1234 said:
ok I am quite confused as i keep getting 0.56 yet the answer is 0.49 m

ill show the teacher my work tom. thank you.

Oh wait, I am sorry, I read the question with the two charges being positive (andyet drew the diagram with + and - :confused:)

BUT the second diagram should be this.


+ ________(0.36m)_________ - ____(r m)_____(P)

SO what you need to do is find the E field at point P due to the -ve charge. Then find the E field due to the +ve charge, and then equate those two.
The distance of the +ve charge to P is (0.36+r) NOT (0.36-r) which would mean that E=0 between the two charges.

All you need to do to rectify the problem is change the sign in your first equation in your working and you should get the correct answer. Sorry again for confusing you there.
 
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