Question

A nonconducting sphere has radius *R* = 2.54 cm and
uniformly distributed charge *q* = +4.89 fC. Take the
electric potential at the sphere's center to be
*V*_{0} = 0. What is *V* at radial distance
from the center **(a)** *r* = 1.50 cm and
**(b)** *r* = *R*? (*Hint*: See
an expression for the electric field.)

Answer #1

We know that electric potential difference between two points is given by:

Now electric field at r distance from center of nonconducting sphere is given by:

E(r) = k*q*r/R^3, So

Using given values: r = 1.50 cm = 0.0150 m

R = 2.54 cm = 0.0254 m

q = 4.89 fC = 4.89*10^-15 C

So,

V(r) = -9*10^9*4.89*10^-15*0.0150^2/(2*0.0254^3)

**V(r) = -3.02*10^-4 V**

Part B

when r = R then:

Since R = 2.54 cm = 0.0254 m, So

V(r) = -9*10^9*4.89*10^-15/(2*0.0254)

**V(r) = 8.66*10-4 V**

**Let me know if you've any query.**

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