Electric Potential from a uniformly charged sphere

In summary, we have a nonconducting sphere with a radius of 2.70 cm and a uniformly distributed charge of +7.00 fC. The electric potential at the center of the sphere is 0. To find the electric potential at a distance r = 1.45 cm, we need to use Gauss' theorem and consider the electric field both inside and outside the sphere. This leads us to the equation V(r) = -int[1/4πεο*Q/r^2*dr] from infinity to R - int[1/4πεο*Qr/R^3*dr] from R to r.
  • #1
Oijl
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0

Homework Statement


A nonconducting sphere has radius R = 2.70 cm and uniformly distributed charge q = +7.00 fC. Take the electric potential at the sphere's center to be V0 = 0.

(a) What is V at radial distance r = 1.45 cm?

(b) What is V at radial distance r = R?

Homework Equations


E = Vdv
V = k (q / r)

The Attempt at a Solution


I was about to just integrate E from zero to r1 and then r2, but then I realized that as r increases, so does q so I can't just have a simple single integration. And then I didn't know what to do. Help? Thanks.
 
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  • #2
Charge density in the sphere = ρ = Q/[4/3*π*R^3]
Charge enclosed in the sphere of radius r = ρ* volume of the sphere of radius r
Q' = { Q/[4/3*π*R^3]}*4/3*π*r^3
= Q*r^3/R^3
Using Gauss theorem, if the electric field E at a distance r is E, then
4πr^2E = Q/εο*r^3/R^3
Or E = 1/4πεο*Qr/R^3 = - dV/dr. Now find the integration.
 
  • #3
Aaaaaahhhhhhhhhhhhh. Thank you so much! I do particularly thank you because you helped me recognize that I need to do much more variable rewriting than I've been doing.
 
  • #4
i had a problem in this question,
i got E = 1/4πεο*Qr/R^3 (using gauss law)
when i applied - dV/dr. , i could not got the answer,I used the basic defination of electric potential that said bring charge from infinity to that pt , i integrated it (-E.dr) from infinity to r,as evident i gt an infinite term in numerator ,please help ?
 
  • #5
I used the basic defination of electric potential that said bring charge from infinity to that pt , i integrated it (-E.dr) from infinity to r,as evident i gt an infinite term in numerator ,please help ?

To find the potential at r, you have to consider the electric field outside and inside the sphere separately.
So V(r) = -int[1/4πεο*Q/r^2*dr] from infinity to R - int[1/4πεο*Qr/R^3*dr] from R to r.
 
  • #6
thanks :) but can you explain in detail that why we follow this approach and what's wrong with d other one?
 
  • #7
Which one is the other approach?
You have tried to find the potential at r using the same expression for E from infinity to r. But it is wrong. Out side the sphere E = 1/4πεο*Q/r^2 and inside the sphere E = 1/4πεο*Qr/R^3. Using these expression find the integration to find the potential at r.
 

Related to Electric Potential from a uniformly charged sphere

1. What is electric potential?

Electric potential is the amount of work needed to move a unit charge from one point to another in an electric field. It is measured in volts (V).

2. How is electric potential calculated?

Electric potential is calculated by dividing the electric potential energy by the charge at a specific point in an electric field. It can also be calculated by multiplying the electric field strength by the distance between the two points.

3. What is a uniformly charged sphere?

A uniformly charged sphere is an object with a spherical shape that has an equal amount of charge distributed evenly over its surface. This means that the electric field strength is the same at every point on the surface of the sphere.

4. How is electric potential affected by a uniformly charged sphere?

The electric potential from a uniformly charged sphere decreases as the distance from the sphere increases. This is because the electric field strength also decreases with distance, causing the potential to decrease as well.

5. What is the relationship between electric potential and electric field strength?

The relationship between electric potential and electric field strength is that the electric field strength is the negative gradient of the electric potential. In other words, the electric field points in the direction of decreasing potential and the magnitude of the field is proportional to the rate of change of potential.

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