Calculating voltage within and outside of a solid sphere

In summary, the sphere has a charge of 3 nC distributed uniformly throughout its volume. The potential difference between a location on the sphere’s surface and the location at the sphere’s centre is calculated to be kq1q2/r. The potential difference between a location on the sphere’s surface and the location at infinity is negligible.
  • #1
starstruck_
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8

Homework Statement


A solid sphere with radius R=12 m has charge Q=3 nC distributed uniformly throughout its volume.
(a) Calculate the potential difference between a location at infinity and a location on the sphere’s surface.
(b) Calculate the potential difference between a location on the sphere’s surface and the location at the sphere’s centre.

Homework Equations


V= ∫ E.R (both vectors)

V= kq1q2/r

The Attempt at a Solution



I'm struggling with understanding this unit in general (first year physics- teacher in high school did not teach electricity) so if I could get some help with this question, it might make the unit in general much clearer for me.

This is what I think:

For part A, I'm integrating between the surface of the sphere and infinity - the force at a distance of infinity is negligible so I'm going to assume 0 for it (prof says it's okay to do so).

so if I'm integrating between infinity and the surface, my integral's upper limit would be ∞ and the lower limit would be R (radius of the sphere).

Since Q is distributed evenly around the volume, I would divide Q by the volume at R; so

R∫k3Q/4πR3? I'm not sure if this is right or let alone whether I'm on the right track.

As for Part B, I would do the same thing but between R and r? Volume would be a factor as the gaussian spheres inside the sphere would have a different amount of charge depending on their volume (as the radius increases) ??
 
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  • #2
starstruck_ said:
Since Q is distributed evenly around the volume, I would divide Q by the volume at R; so

R∫k3Q/4πR3?
You need to understand what you are effectively calculating there.
Dividing by the volume gives the charge density inside the sphere. This is a constant. Your integral is undefined because you have not specified the variable of integration (the "dx"). It cannot be R because R is a constant.
Anyway, there does not seem to be any value in integrating the charge density within the sphere over a range outside it.

What do you know about the field outside a uniformly charged spherical shell?
 

1. How do you calculate the voltage within a solid sphere?

The voltage within a solid sphere can be calculated using the formula V = kQ/r, where V is the voltage, k is Coulomb's constant, Q is the charge of the sphere, and r is the radius of the sphere.

2. What is the equation for calculating voltage outside of a solid sphere?

The equation for calculating voltage outside of a solid sphere is V = kQ/r, where V is the voltage, k is Coulomb's constant, Q is the charge of the sphere, and r is the distance from the center of the sphere.

3. Can the voltage within a solid sphere be negative?

Yes, the voltage within a solid sphere can be negative if the charge within the sphere is negative. This indicates that the electric field within the sphere is directed towards the center.

4. How does the voltage change as you move from the center to the surface of a solid sphere?

The voltage decreases as you move from the center to the surface of a solid sphere. This is because the distance from the center increases, resulting in a smaller value for r in the voltage equation.

5. How does the voltage outside of a solid sphere vary with distance?

The voltage outside of a solid sphere follows an inverse relationship with distance. As the distance from the center of the sphere increases, the voltage decreases because the electric field becomes weaker at larger distances.

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