How Do You Calculate the Potential Difference in a Charged Spherical Insulator?

[deerhunt713]

A solid insulating sphere has radius a = 3.2 meters and total charge Q = 3.6 Coulomb's. Calculate the potential difference between the center of the sphere and a point r = 0.64 meters from the center of the sphere. (A positive answer represents the center is at a lower potential, a negative answer corresponds to the center being at a higher potential.)



a similar problem was solved for a cylinder as an example, but i am unable to figure out how to solve for a sphere. please help

 

[Kalvarin]

Use gauss's law to work out the electric field produced by the sphere as the sphere has symmetry. The field inside an insulating sphere is the same as the field outside it so we don't have to worry about the fields being different, the field we get from gauss's law is a general result. Now all you need to calculate is the negative line integral of the electric field from the point outside the sphere to the point at the centre. This is the potential difference between the outside point and the centre.

 

[deerhunt713]

im still having trouble/ not completely grasping the concepts. pls help its due in the morning

 

[Kalvarin]

\oint E.da = Qenc/Eo

Where Qenc is the charge enclosed within the surface and Eo is the permitivity of free space.

Because the field has symmetry, in this case spherical, we can simplify the above equation to:

E\ointda = Qenc/Eo

Which becomes:

E A = Qenc/Eo

Where A is the area of the gaussian surface.

So now we put a sphereical gaussian surface around the sphere so that it is totally enclosed. So A = 4\pir^2, the area of a sphere.

So we get the equation for the electric field to be:

E = Qenc/4\piEo r^2

This holds for any straight line directed radially outward from the centre of the sphere because the sphere has sphereical symmetry.

Now the potential is given by:

V = -\intE.dl

You need to evaulate it from 0.64 to 0 to find the potential difference. Will leave that part to you :)