Magnetic fields

  • Thread starter Matt1234
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  • #1
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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 im looking for r, but the way i setup the equation gives me 2 unknowns.
 

Answers and Replies

  • #2
rock.freak667
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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.
 
  • #3
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im sorry i dont 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. Im not 100% sure about this part "put the point where E=0 as a distance x from one charge"
 
  • #4
rock.freak667
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i tried that and i get 0.2 which is wrong. Im 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 '??'
 
  • #5
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(0.36 - r) i believe is what im 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:
  • #6
rock.freak667
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(0.36 - r) i believe is what im looking for?

Making this question quadratic?

Yes.
 
  • #7
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ok im 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:
  • #8
rock.freak667
Homework Helper
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ok im 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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