Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply that of a point chargeQ. The electric field Ein inside the sphere (r≤R) is radially outward with field strength
Ein=14πϵ0QR3r

Q: a) The electric potential Vout outside the sphere is that of a point charge Q. Find an expression for the electric potential Vin at position r inside the sphere. As a reference, let Vin=Vout at the surface of the sphere.

Express your answer in terms of the given quantities and appropriate constants.

b) What is the ratio VcenterVsurface?

a: 2/3
b: 2
c: 1/2
d: 3/2
e: 1
c) Graph V versus r for 0≤r≤3R. [x axis: r(R)y axis:V((1/4piE0)(Q/R))]

a) Vb - Vc = K Q / 2r³ (R² –r²) ,   b)  Vc / Vs = ½   The correct answer is c

Explanation:

The electric potential and the electric field are related by the expression

      dV = - E .ds

In this case the electric field lines are radial and normal to the sphere is also radial, so the angle between them is zero and the scalar product is reduced to the algebraic product.

a) For potential outside the sphere

     ∫ dV = - ∫ E dr

    Vb - Va = - k Q ∫ dr / r²

   Vb - Va = k Q (1 / rb - / ra)

If we take the power to be zero for the infinite distance

    Vb = k Q  1 / rb

Now let's calculate the potential for a point inside the sphere

         E = k Q/R³   r

         k = 1 / 4π eo

        Vb-Vc = - k Q / R³ ∫ r dr

        Vb - Vc = - k Q / R³   r² / 2

We evaluate for lower limit with R and upper limit with r

       Vb - Vc = K Q / 2r³ (R² –r²)

b) the relation of the potential in the center (r = 0) and the surface (r = R)

       Vc = k Q / 2R³ R² = k Q / 2R

       Vs = k Q / R

The relationship between these potentials is

        Vc / Vs = ½

  The correct answer is c

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