Electric potential at point x on the axis of a ring of charge density eta

In summary, to find the electric potential at point x on the axis of a ring of charge density \eta = cr, you can integrate the charge density over the disk to compute the total charge, Q, and then use this to eliminate c in the equation for electric potential. The final expression should include R and Q, but not c.
  • #1
Bng1290
5
0
Electric potential at point x on the axis of a ring of charge density "eta"

Homework Statement


A circular disk of radius R and total charge Q has the charge distributed with surface charge density [tex]\eta[/tex] = cr, where c is a constant. Find an expression for the electric potential at distance z on the axis of the disk. Your expression should include R and Q, but not c.

Homework Equations



[tex]\eta[/tex]=cr where c is constant
V=(1/4pi[tex]\epsilon[/tex])(Q/r)
V=[tex]\Sigma[/tex]Vi

The Attempt at a Solution


So what I did was to sum all Vi and i was able to pull (1/4pi[tex]\epsilon[/tex]) and (1/sqrt(z^2+R^2) out which leaves me with Q left in the sum which I know i need to relate to [tex]\eta[/tex] in some way. The problem I'm having here is that I just don't understand how to work with [tex]\eta[/tex]=cr in such a way as to get rid of the constant c in my answer.

I feel like I'm not grasping this problem as a whole so any help would be wonderful. Thanks!
 
Physics news on Phys.org
  • #2


To eliminate c, since \eta = cr, you can integrate the charge density over the disk to compute the total charge, Q. This should give you c in terms of Q and R.
 
  • #3


Nice! Thanks so much!
 

1. What is the formula for calculating electric potential at point x on the axis of a ring of charge density eta?

The formula for calculating electric potential at point x on the axis of a ring of charge density eta is V = k*Q*x/(x^2 + R^2)^(3/2), where k is the Coulomb constant, Q is the total charge of the ring, x is the distance from the center of the ring to the point, and R is the radius of the ring.

2. How does the electric potential at point x change as the distance from the ring changes?

The electric potential at point x on the axis of a ring of charge density eta decreases as the distance from the ring increases. This is because the electric potential is inversely proportional to the distance from the source of the electric field, which in this case is the ring.

3. Can the electric potential at point x ever be negative?

Yes, the electric potential at point x can be negative. This occurs when the point is located within the ring, at a distance less than the radius. In this case, the electric potential is negative because the electric field is in the opposite direction of the direction of the point's movement.

4. How does the charge density eta affect the electric potential at point x?

The charge density eta has a direct effect on the electric potential at point x. As the charge density increases, the electric potential also increases. This is because a higher charge density means a greater amount of charge is present in the ring, resulting in a stronger electric field and higher electric potential at point x.

5. Is the electric potential at point x affected by the size of the ring?

Yes, the size of the ring, specifically the radius, does affect the electric potential at point x. As the radius increases, the electric potential at point x decreases. This is because a larger radius means a greater distance between the point and the ring, resulting in a weaker electric field and lower electric potential at point x.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
351
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
992
  • Introductory Physics Homework Help
Replies
4
Views
2K
Replies
1
Views
123
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
208
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
870
Back
Top