Is My Equation for Finding Electric Potential in the Correct Form?

In summary, the equation used to solve for the total electric potential is incorrect because the 'z' term is incorrect. The equation should instead be integrated over the range x = r to x = R, corresponding to the radial extremes of the disk.
  • #1
VitaX
184
0

Homework Statement



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Homework Equations



V = (sigma)/(2*epsilon_0)*((z^2 + R^2)^(1/2) - z)

The Attempt at a Solution



This was my full equation to find the total electric potential:

V = (sigma)/(2*epsilon_0)*((z^2 + R^2)^(1/2) - z) - (sigma)/(2*epsilon_0)*((z^2 + r^2)^(1/2) - z)

But I got the answer wrong. Is this equation even correct here? I'm not seeing what I did wrong if it is the right equation to use.
 
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  • #2
I think that you may want to verify your "Relevant equation". That "-z" term looks suspicious. How did you find this equation?
 
  • #3
gneill said:
I think that you may want to verify your "Relevant equation". That "-z" term looks suspicious. How did you find this equation?

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  • #4
Ah. The 'z' results from the disk going from radius 0 to radius R. Your disk has a minimum radius that is not zero, so rather than a 'z' term there should be another sqrt(z2 + r2) in its place, where r is the inner radius of the ring.
 
  • #5
gneill said:
Ah. The 'z' results from the disk going from radius 0 to radius R. Your disk has a minimum radius that is not zero, so rather than a 'z' term there should be another sqrt(z2 + r2) in its place, where r is the inner radius of the ring.

I see, so if I do that it will complete the equation needed to solve this?
 
  • #6
That should do it, but it might be preferable for you to do the integration and confirm where the term comes from. You might have to do something similar in an exam at some point.
 
  • #7
True, I know I start with dV = (1/4*pi*e_0) \integral (dQ/r)

Next step would be to say dQ = sigma*dx where sigma = Q/Area. Do I use Area = 4*pi*R^2 here? I'm a bit confused as to how about going to set this up and where the R and r come into play.
 
  • #8
Yes, it gets a bit confusing because there are so many different r's floating around in this problem. Rename some. Call the distance from a given ring of charge to the point on the z-axis d. That takes the place of the r in the \integral (dQ/r), and is given by sqrt(z2 + x2), where here x is the radius of the given differential ring of charge. The integration limits will be x = r to x = R, corresponding to the radial extremes of the disk.

Have a look http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potlin.html" for the geometry and setup of the integral.
 
Last edited by a moderator:

1. What is electric potential?

Electric potential is a measure of the potential energy of a charged particle in an electric field. It represents the amount of work required to move a unit charge from one point to another in an electric field.

2. How is electric potential different from electric field?

Electric field is a measure of the force experienced by a charged particle in an electric field, whereas electric potential is a measure of the energy of the particle in the field. Electric potential is related to electric field by a simple mathematical formula.

3. How do you find the electric potential at a point?

The electric potential at a point can be found by dividing the work done in moving a unit charge from infinity to that point by the magnitude of the charge. This can be calculated using the formula V = W/q, where V is the electric potential, W is the work done, and q is the magnitude of the charge.

4. What is the unit of electric potential?

The unit of electric potential is volts (V), which is equivalent to joules per coulomb (J/C).

5. Can electric potential be negative?

Yes, electric potential can be negative. A negative potential indicates that work must be done against the electric field in order to move a unit charge from infinity to that point. This means that the electric field is opposing the motion of the charged particle.

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