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Potential difference on a charge between insulating plates

  • Thread starter clope023
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  • #1
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Homework Statement



A small sphere with mass 1.50g hangs by a thread between two parallel vertical plates 5cm apart. THe plates are insulating ahnd have uniform surface charge densities + and - [tex]\sigma[/tex]. The charge on the shpere is q = 8.90 x 10^-6 C. What potential difference between the plates will cause the the thread to assume an angle of 30 degrees with the vertical?

Homework Equations



Va->b = Ex

E = [tex]\sigma[/tex]/[tex]\epsilon[/tex]0

Va->b = [tex]\sigma[/tex]d/[tex]\epsilon[/tex]0


The Attempt at a Solution



I am actually a little stuck as to where to begin, I know the sphere orginally hangs on its string at x = .025m, and since the opposing plates are oppositely charged the positive sphere is attracted to the negative plate and raised a certain distance at an angle of 30 degrees.

would I be right in assuming I could find the magnitude of the surface charge density utilizing the potential with formula kq/r, with r = .025m; then use that density in the equation with potential at r = .025cos30 (the point where the shpere is pulled by the negative plate)?

so it would turn out like this:

[tex]\sigma[/tex] = [tex]\epsilon[/tex]0V.025->0/.05

then use this value in the equation:

Va->b = [tex]\sigma[/tex].025cos30/V.025->0

?

any help is greatly appreciated.
 

Answers and Replies

  • #2
Doc Al
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I am actually a little stuck as to where to begin, I know the sphere orginally hangs on its string at x = .025m, and since the opposing plates are oppositely charged the positive sphere is attracted to the negative plate and raised a certain distance at an angle of 30 degrees.
Did you draw a free body diagram showing all the forces acting on the sphere? What must the net force be in all directions?

would I be right in assuming I could find the magnitude of the surface charge density utilizing the potential with formula kq/r, with r = .025m; then use that density in the equation with potential at r = .025cos30 (the point where the shpere is pulled by the negative plate)?
No. That formula is for the potential from a point charge--not relevant here.

Hint: Find the value of the electric field' between the plates.
 
  • #3
987
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Did you draw a free body diagram showing all the forces acting on the sphere? What must the net force be in all directions?


No. That formula is for the potential from a point charge--not relevant here.

Hint: Find the value of the electric field' between the plates.
I'm not sure what you mean by net force in all direction, the charge is positive so it's electric field is going outward as is the field of the positive plate, the negative field's plate is going inward and thus attracting the outward field of the point charge on the string.

about your hint, didn't I do that already? E = sigma/eplison0
 
  • #4
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First of all, your use of LateX is really annoying, sorry :(
You don't have to type each letter in a seperate tex block, just type them all in one block.
Also, fractions can be made using: \frac{a}{b} = [tex]\frac{a}{b}[/tex].

On to your question...

I don't see you using any force in your attempt. Surely the only thing that can move the sphere will be a force?
The force is a result of the positive plate charge attracting the negative sphere; this is what will cause the sphere to move.

If you can figure out what force is needed to make the sphere hang at an angle of 30 degrees, and you can also express this force in terms of the two charges, then you can also find the potential difference that will cause the force you are looking for.
 
  • #5
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sorry about that, oh and the sphere is positive.

would this be correct as far as the force goes?

F = qE = qsigma/eplison0
 
  • #6
Doc Al
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I'm not sure what you mean by net force in all direction, the charge is positive so it's electric field is going outward as is the field of the positive plate, the negative field's plate is going inward and thus attracting the outward field of the point charge on the string.
My point is that the sphere is in equilibrium. What does that tell you?

You must begin by identifying all the forces acting on the sphere. (I count three.)

about your hint, didn't I do that already? E = sigma/eplison0
This problem is oddly worded. I assume that the statement about surface charge was a mistake. (If you really had the surface charge, then you'd know the field. If you knew the field, you can calculate the potential difference without worrying about all the other info in the problem.)

In any case, one of those three forces acting on the sphere is the electrical force. Use the conditions for equilibrium to solve for the E field, then calculate the potential difference.
 
  • #7
Doc Al
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would this be correct as far as the force goes?

F = qE = qsigma/eplison0
That's the force due to the electric field. (You can skip the last term, unless a later part of the problem asks you to calculate the surface charge density.)
 
  • #8
987
123
My point is that the sphere is in equilibrium. What does that tell you?

You must begin by identifying all the forces acting on the sphere. (I count three.)


This problem is oddly worded. I assume that the statement about surface charge was a mistake. (If you really had the surface charge, then you'd know the field. If you knew the field, you can calculate the potential difference without worrying about all the other info in the problem.)

In any case, one of those three forces acting on the sphere is the electrical force. Use the conditions for equilibrium to solve for the E field, then calculate the potential difference.
the field is uniform inside the plates and perpendicular to their area, so by Guass' law Qencl = sigmaA and the total flux integral inside the plates is EA, so by Gauss' law EA = sigmaA/epsilon0 and E = sigma/eplison0.

nope, that's how the book worded it, the surface charge's value was not given, they just said that there was one on each plate.

saw from your other answer that I got the force due to field, are you saying I can calculate the work done by the force to find the potential?
 
  • #9
Doc Al
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the field is uniform inside the plates and perpendicular to their area, so by Guass' law Qencl = sigmaA and the total flux integral inside the plates is EA, so by Gauss' law EA = sigmaA/epsilon0 and E = sigma/eplison0.
That relationship for E in terms of surface charge density is true, but not needed here.

nope, that's how the book worded it, the surface charge's value was not given, they just said that there was one on each plate.
Seems just an odd side comment to make. (Kind of like saying: A ball drops a distance d. How far does it fall? Well... d!) We know it has some surface charge.

saw from your other answer that I got the force due to field, are you saying I can calculate the work done by the force to find the potential?
You certainly could. The field is constant between the plates. What's the relationship between potential difference and electric field?
 

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