Question

There is a uniformly charged non conducting solid sphere made of material of dielectric constant one. If electric potential at infinity be zero, then the potential at its surface is V. If we take electric potential at its surface to be zero, then the potential at the centre will be (1) 3V/2 (2) V/2 (3) V (4) Zero

Answer 1

The expression for the electric potential at the surface (referenced to 0 at infinity) is pretty straightforward, since the entire spherically symmetric distribution is contained within that position.  Therefore, it presents the same electric potential at that position as a point charge would, with all the charge of the sphere collapsed to its center.  Thus, we know that:

 

V = kQ/R  

 

where 

 

R = radius of the sphere

Q = total charge on the sphere

k = Coulomb's constant

 

Now. to determine the potential difference between the surface and the center, we can use:

 

Vab = - ∫ab E · ds

 

If I use r as my radius coordinate, and choose the center of the sphere as r = 0, I need to determine an expression for E(r) inside the sphere.  Since the sphere is uniform, the electric field at any position r away from the center is the electric field generated by the charge contained within a (conceptual) sphere with that radius, which acts like a point charge with all the charge of that enclosed sphere collapsed to the center (this is a result of Gauss' law).  The charge outside of r does not contribute.

 

Therefore:

 

E(r) = kqenc/r2

 

where qenc is the charge enclosed with a sphere of radius r.  Since the distribution is uniform, this will just be a direct proportion of the total charge, proportional to how much of the total volume of the full sphere is enclosed within a sphere of radius r.

 

qenc = Q(Venc/Vtot)

 

This leads to:

 

E(r) = kQr/R3

 

Since it is not specified whether the charge on the sphere is positive or negative, we do not know if the electric field points out from the center, or in toward the center.  I will assume positive for the purpose of this problem.  If it is negative, only the sign -- but not the form -- of the answer will change.

 

The integral for potential difference between the surface and the center becomes:

 

Vsc = - ∫R0 (kQr/R3)dr

 

This is because the electric field (from a symmetry argument) is only radial, so only the steps I take from the surface to the center along the radial direction count.  Also, since I defined the center as r = 0 and the surface as r = R, I defined the outward direction as positive, and (with my positive charge assumption) the electric field also points outward.  So the cos(theta) term in the dot product becomes just cos(0) = 1.  This results in the integral above

 

Integrating that gives:

Vab = kQ/2R

 

This is a positive difference traveling from the surface to the center, which jibes with my positive charge definition for the sphere.  Electric fields point from positive to negative charges, and -- since electric potential is defined to be high near positive charges, and low near negative charges -- electric fields always point from high to low potential.  Hence, going from the surface to the center -- moving opposite the electric field -- should result in an increase in potential.

 

If I relate Vab to the definition of V above, I get:

 

Vab = V/2

 

This means the center of the sphere is higher in potential that the surface by a factor of V/2.  So, if the surface is defined to be V, the potential at the center must be V + V/2.

 

This same result holds, in its own way, if the charge is negative in the sphere.  In this case, the center is V/2 lower than the surface in potential... but the surface potential is negative to begin with, resulting the in the same relationship, only with a negative sign in front of it.

V/2

Answer 2

Explanation:

hope you understand this answer