Question

A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the sphere's center to be V0 = 0. What is V at radial distance from the center (a) r = 1.60 cm and (b) r = R? (Hint: See an expression for the electric field.)

Answer 1

(a). The electric potential at 1.650 cm is
`-1.219*10^(-4)\ V`.

(b). The electric potential at 2.81 cm is
`-3.759*10^(-4)\ V`.

Step-by-step explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

`V = 0`

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

`V_(r)-V_(0)=-\int_(0)^(r){E(r)}dr`

`V_(r)-0=-\int_(0)^(r){(qr)/(4\pi\epsilon_(0)R^3)}dr`

`V_(r)=-((qr^2)/(8\pi\epsilon_(0)R^3))_(0)^{1.60*10^(-2)}`

`V_(r)=-((2.35*10^(-15)*(1.60*10^(-2))^2)/(8*\pi*8.85*10^(-12)*(2.81*10^(-2))^3))`

`V_(r)=-0.00012190\ V`

`V_(r)=-1.219*10^(-4)\ V`

The electric potential at 1.650 cm is
`-1.219*10^(-4)\ V`.

(b). We need to calculate the potential at a distance r = R

Using formula of potential difference

`V_(R)=-(2.35*10^(-15))/(8\pi*8.85*10^(-12)*2.81*10^(-2))`

`V_(R)=-0.0003759\ V`

`V_(R)=-3.759*10^(-4)\ V`

The electric potential at 2.81 cm is
`-3.759*10^(-4)\ V`.

Hence, This is the required solution.