Electric Field between two Charges

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  • #1
Dillion
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Homework Statement


A positive charge of 3 microCouloumbs is at the origin. A negative charge with the same magnitude is 8 m away along the x direction.
a) What are the magnitude and direction of the electric field at the points between the two charges, 2, and 4, and 6m from the positive charge?
b) What is the electric potential at each of these three points? (The reference point, where V=0 is infinitely far away).

Homework Equations


V=Ed

Electric Field = Qk/r^2

The Attempt at a Solution


a) (3*10^-6)(9*10^9)/2^2 = 6750
(3*10^-6)(9*10^9)/6^2 = 750
7500 to the right
(3*10^-6)(9*10^9)/4^2 = 3375 to the right


(3*10^-6)(9*10^9)/6^2 = 750
(3*10^-6)(9*10^9)/2^2 = 6750
7500 to the right

b) V=Ed
Electric field at 2m
7500 * 2 = 15000

Electric field at 4 m
3375 * 4 = 13500

Electric field at 6 m
7500 * 6 = 45000


I'm confident in A but not in b. Please help
 

Answers and Replies

  • #2
gneill
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For part b, you can see from part a that the electric field is not uniform (same magnitude and direction) as you move from the origin towards the second charge. So your equation V = E*d doesn't hold because E is not a constant value. You would have to integrate: ∫E(x)⋅dx.

Instead, use the formula for the electric potential vs distance for a point charge.
 
  • #3
Dillion
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For part b, you can see from part a that the electric field is not uniform (same magnitude and direction) as you move from the origin towards the second charge. So your equation V = E*d doesn't hold because E is not a constant value. You would have to integrate: ∫E(x)⋅dx.

Instead, use the formula for the electric potential vs distance for a point charge.

What is that formula?
 
  • #4
gneill
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What is that formula?
It should be in your text or course notes. It's a very common formula based on Coulomb's Law. You could also do a web search on "Electric potential due to a point charge".
 
  • #5
Dillion
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It should be in your text or course notes. It's a very common formula based on Coulomb's Law. You could also do a web search on "Electric potential due to a point charge".

Electric Potential = qsk/r ?
 
  • #6
gneill
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Electric Potential = qsk/r ?
Hmm. Maybe... Can you explain what "qsk" represents?
 
  • #7
Dillion
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point source charge
 
  • #8
gneill
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point source charge
I'd have to say almost but not quite. Is "qsk" one variable? Or is Coulomb's constant "k" a separate value?
 
  • #9
Dillion
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I'd have to say almost but not quite. Is "qsk" one variable? Or is Coulomb's constant "k" a separate value?
oh yeah sorry qs times k which is 9*10^9 all divided by r
 
  • #10
gneill
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oh yeah sorry qs times k which is 9*10^9 all divided by r
Okay, presumably qs is meant to be qs. (Note that you can use the x2 and x2 buttons in the edit panel header to create sub- and superscripts).

So, using that formula in much the same way as you used the one for the electric field, find the electric potential at the given locations. Note that electric potential is a scalar quantity, so you don't have to worry about vector addition this time.
 
  • #11
Dillion
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Okay, presumably qs is meant to be qs. (Note that you can use the x2 and x2 buttons in the edit panel header to create sub- and superscripts).

So, using that formula in much the same way as you used the one for the electric field, find the electric potential at the given locations. Note that electric potential is a scalar quantity, so you don't have to worry about vector addition this time.

b) V = qsk/r
(9*10^9)(3*10^-6)/2 = 13500
(9*10^9)(3*10^-6)/6 = 4500
= 18000

is this how I would do find the electric potential at point 2?
 
  • #12
gneill
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Very close. You need to take into account the signs of the charges. Isn't point two equidistant from both charges? Point one was at 2 m from the positive charge, point two was at 4 m, and point three at 6 m?
 
  • #13
Dillion
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Very close. You need to take into account the signs of the charges. Isn't point two equidistant from both charges? Point one was at 2 m from the positive charge, point two was at 4 m, and point three at 6 m?

Wouldn't the signs all be positive because a positive particle would move to the right away from the origin and towards the negative particle at 8 m
 
  • #14
gneill
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Wouldn't the signs all be positive because a positive particle would move to the right away from the origin and towards the negative particle at 8 m
Nope. As I said before, electric potential is a scalar property, not a vector. All that matters is the sign of the charge and the distance.
 
  • #15
Dillion
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Nope. As I said before, electric potential is a scalar property, not a vector. All that matters is the sign of the charge and the distance.

I'm not sure how to continue. Having a positive charge near the origin would increase electric potential and if it was closer to the negative charge electric potential would decrease
 
  • #16
gneill
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I'm not sure how to continue. Having a positive charge near the origin would increase electric potential and if it was closer to the negative charge electric potential would decrease
Yes. That's not a problem.

Edit: If you were to plot the electric potential vs position between the two charges you'd get a curve that looks something like this:

upload_2016-4-12_23-45-7.png
 
Last edited:
  • #17
Dillion
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Yes. That's not a problem.

Edit: If you were to plot the electric potential vs position between the two charges you'd get a curve that looks something like this:

View attachment 99012

Need more help please
 
  • #18
gneill
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Apply the potential formula at each point using the signed charges and distances from each charge. Sum the contributions from each.
 
  • #19
Dillion
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Apply the potential formula at each point using the signed charges and distances from each charge. Sum the contributions from each.

b) V = qsk/r
(9*10^9)(3*10^-6)/2 = 13500
(9*10^9)(3*10^-6)/6 = 4500
= 18000

I still don't understand why this is incorrect. Everything is in the positive direction
 
  • #20
gneill
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I still don't understand why this is incorrect. Everything is in the positive direction
You aren't paying attention. You must incorporate the signs of the charges. Direction does not matter: voltage is a scalar quantity and has no direction associated with it. All that matters is the signed charge and the distance from it.
 
  • #21
Dillion
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You aren't paying attention. You must incorporate the signs of the charges. Direction does not matter: voltage is a scalar quantity and has no direction associated with it. All that matters is the signed charge and the distance from it.
b) V = qsk/r
(9*10^9)(3*10^-6)/2 = 13500
(9*10^9)(3*10^-6)/6 = -4500 (electric potential decreases)
-4500+13500 = 9000

?
 
  • #22
gneill
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Yup. But you should incorporate the sign of the charge into the equation so it's obvious where the sign of the result comes from. Thus:

(9*10^9)(-3*10^-6)/6 = -4500 (electric potential decreases)
 

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