Question

A nonconducting sphere has radius R = 2.31 cmand uniformly distributed charge q = +3.50 fC. Take the electric potential at the sphere’s center to be V = 0 . What is Vat radial distance (a) r = 1.45 cmand (b) r = R. (Hint: See Module 23-6.)

Short answer

1. The value of electric potential is $V\left(r\right)=-2.68\times {10}^{-4}V$.
2. The value of electric potential is $V\left(R\right)=-6.81\times {10}^{-4}V$.

Step-by-step solution

Given data:

• Radius of a nonconducting sphere is $R=2.31cm=0.0231m$.
• Charge distributed on the nonconducting sphere is $q=+3.50fC=+3.50\times {10}^{-15}C$
• Electric potential at the sphere’s center is V (0) = 0.
• The radial distance,$r=1.45cm=0.0145m$

Understanding the concept

The electric potential V at the surface of a drop of charge q and radius R is given by,

$\begin{array}{rcl}\mathit{V}& \mathbf{=}& \frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{R}}\\ & \mathbf{=}& \frac{\mathbf{k}\mathbf{q}}{\mathbf{R}}\end{array}$

Here, k is the Coulomb’s constant having a value as below.

$\begin{array}{rcl}k& =& \frac{1}{4\pi {\epsilon }_0}\\ & =& 8.99\times {10}^{9}N·m^2/C^2\end{array}$

(a) Calculate V at radial distance r = 1.45 cm :

The potential as a function of r is,

$\begin{array}{rcl}V\left(r\right)& =& V\left(0\right)-\underset{0}{\overset{r}{\int }}E\left(r\right)dr\\ & =& 0-\underset{0}{\overset{r}{\int }}\frac{qr}{4\pi {\epsilon }_0{R}^3}dr\\ & =& -\frac{1}{2}\frac{q{r}^2}{4\pi {\epsilon }_0{R}^3}\end{array}$

Substitute known values in the above equation.

role="math" localid="1662549529954" $\begin{array}{rcl}V\left(r\right)& =& -\frac{1}{2}\frac{\left(+3.50\times {10}^{-15}C\right){\left(0.0145m\right)}^2\left(8.99\times {10}^{9}N·m^2/C^2\right)}{{\left(0.0231m\right)}^3}\\ & =& -2.68\times {10}^{-4}V\end{array}$

Hence, the value of electric potential is $\begin{array}{rcl}& & -2.68\times {10}^{-4}V\end{array}$.

(b) Calculate V at radial distance r = R :

Since the potential difference is,

$\begin{array}{rcl}∆V& =& \frac{q}{8\pi {\epsilon }_{0}R}\\ V\left(0\right)-V\left(R\right)& =& \frac{q}{8\pi {\epsilon }_{0}R}\\ 0-V\left(R\right)& =& \frac{q}{8\pi {\epsilon }_{0}R}\end{array}$

Substitute known values in the above equation.

role="math" localid="1662550241996" $\begin{array}{rcl}V\left(R\right)& =& -\frac{\left(8.99\times {10}^{9}N.m^2/C^2\right)\left(3.50\times {10}^{-15}C\right)}{2{\left(0.0231m\right)}^3}\\ & =& -6.81\times {10}^{-4}V\end{array}$

Hence, the value of electric potential is $\begin{array}{rcl}& & -6.81\times {10}^{-4}V\end{array}$.