Radii of eqiuipotential surfaces of a charged metal sphere

In summary, the problem involves a metal sphere with a radius of 0.29 m and a charge of 0.90 µC. The task is to determine the radius of the first, tenth, and 100th equipotential surfaces from the surface of the sphere, with 100 V intervals. Using the equation V=kQ/r, the potential at the surface of the sphere is found to be 27931.03V. To find the radius of each equipotential, 100 V is added to the potential for the first, 1000 V for the tenth, and 10000 V for the 100th, but the results were incorrect. The radius of the sphere itself may also need to be taken
  • #1
RoosterCancer
3
0

Homework Statement



A metal sphere of radius r0 = 0.29 m carries a charge Q = 0.90 µC. Equipotential surfaces are to be drawn for 100 V intervals outside the sphere. Determine the radius r of the following equipotentials from the surface.
a) first:___m
b) tenth:___m
c) 100th:_____m


Homework Equations



V=kQ/r --> r=kQ/V



The Attempt at a Solution



I've tried plugging the numbers from the problem in the equation:

r=(8.99 x 10^9 NM^2/C^2)*(0.9 x 10^-6 C)/(100V) = 80.91m

and it's incorrect. I assume the radius of the sphere itself comes into play, but I'm not sure where. Can anyone help?
 
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  • #2
RoosterCancer said:

Homework Statement



A metal sphere of radius r0 = 0.29 m carries a charge Q = 0.90 µC. Equipotential surfaces are to be drawn for 100 V intervals outside the sphere. Determine the radius r of the following equipotentials from the surface.
a) first:___m
b) tenth:___m
c) 100th:_____m

Homework Equations



V=kQ/r --> r=kQ/V

The Attempt at a Solution



I've tried plugging the numbers from the problem in the equation:

r=(8.99 x 10^9 NM^2/C^2)*(0.9 x 10^-6 C)/(100V) = 80.91m

and it's incorrect. I assume the radius of the sphere itself comes into play, but I'm not sure where. Can anyone help?
What is the potential at the surface of the ball ?
 
  • #3
I was able to find the potential of the ball:

V=(9 x10^9NM^2/C^2)*(0.9 x10^-6 C)/0.29m = 27931.03V

and was able to answer part a, by adding 100 V to the ball's potential and using the formula. So, for b, I added 1000 V to the potential and for c I added 10000 V to the potential, but the results were incorrect.
 

FAQ: Radii of eqiuipotential surfaces of a charged metal sphere

1. What is a charged metal sphere?

A charged metal sphere is a spherical object made of a conductive material, such as metal, that has a net electric charge. This means that the number of positive and negative charges within the sphere is not balanced, resulting in an overall charge.

2. What are equipotential surfaces?

Equipotential surfaces are imaginary surfaces that represent points in space with the same electric potential. This means that any point on an equipotential surface will have the same amount of electric potential energy.

3. How are the radii of equipotential surfaces determined for a charged metal sphere?

The radii of equipotential surfaces for a charged metal sphere are determined by the distance from the center of the sphere to the surface. The potential at each point on the surface is equal, so the radius will vary based on the strength of the charge and the distance from the center.

4. Why do the radii of equipotential surfaces decrease as you move closer to the center of the charged metal sphere?

The radii of equipotential surfaces decrease as you move closer to the center of a charged metal sphere because the electric potential increases as you get closer to the center. This means that the electric field is stronger closer to the center, resulting in a smaller radius for the equipotential surface.

5. How do the radii of equipotential surfaces change when the charge on the metal sphere is increased?

When the charge on the metal sphere is increased, the radii of the equipotential surfaces will decrease. This is because a larger charge results in a stronger electric field, causing the equipotential surfaces to be closer together and have a smaller radius. The opposite is also true - decreasing the charge on the sphere will result in larger radii for the equipotential surfaces.

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