A metal sphere of radius $15\text{cm}$ has a net charge of$3.0\times {10}^{-8}\text{C}$ . (a) What is the electric field at the sphere’s surface? (b) If$V=0$ at infinity, what is the electric potential at the sphere’s surface? (c) At what distance from the sphere’s surface has the electric potential decreased by$\text{500 V}$ ?

The Coulomb’s constant is,

$$\begin{gathered} k=\frac{1}{4\pi {\epsilon }_o} \\ =9.0\times {10}^{9}N.m^2/C^2 \end{gathered}$$

The charge, $q=3.0\times {10}^{-8}\text{C}$

The radius, $R=15\text{cm}=0.15\text{m}$

Electric potential,$V=500\text{V}$

The electric potential due to a collection of point charges is,

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_o}\left(\frac{q}{R}\right) \\ =k\left(\frac{q}{R}\right) \end{gathered}$$

Here, $q$is the charge,$R$is the radius of the metal sphere, and $k$ is the Coulomb’s constant.

The electric field at the surface of the sphere is,

$$\begin{gathered} E=k\left(\frac{q}{R^2}\right) \\ =\left(9.0\times {10}^{9}N.m^2/C^2\right)\frac{\left(3.0\times {10}^{-8}C\right)}{{\left(0.15m\right)}^2} \\ =1.2\times {10}^{4}N/C \\ =12kN/C \end{gathered}$$
Therefore, the electric field at the spheres surface is$12kN/C$

The electric potential at the surface of the sphere is

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_o}\left(\frac{q}{R}\right) \\ =k\left(\frac{q}{R}\right) \end{gathered}$$

Put $2.0\times {10}^{-8}\text{C}$ for $q$ , and $0.15\text{m}$ for $R$ in the above equation.

role="math" localid="1662550507745" $$\begin{gathered} V=\left(9.0\times {10}^9\right)\frac{\left(3.0\times {10}^{-8}\right)}{\left(0.15\right)} \\ =1.8\text{kV} \end{gathered}$$

Therefore, the electric potential at the surface of the sphere's surface is $1.8\text{kV}$.

The expression electric potential at a distance $x$ is,

$V\left(x\right)=\frac{q}{4\pi {\epsilon }_o}\left(\frac{1}{R+x}\right)$

The electric potential of the spherical surface is decreased by

$\Delta V=V\left(x\right)-V$

Substitute $\frac{q}{4\pi {\epsilon }_o}{\left(\frac{1}{R+x}\right)}_{}$for $V{\left(x\right)}_{}$and $\frac{q}{4\pi {\epsilon }_o}\left(\frac{1}{R}\right)$in the above equation.

$$\begin{gathered} \Delta V=\frac{q}{4\pi {\epsilon }_o}\frac{1}{R+x}-\frac{q}{4\pi {\epsilon }_o}\left(\frac{1}{R}\right) \\ =\frac{q}{4\pi {\epsilon }_o}\left(\frac{1}{R+x}-\frac{1}{R}\right) \\ =\frac{q}{4\pi {\epsilon }_o}\left(\frac{R-R-x}{\left(R+x\right)R}\right) \\ =-\frac{qx}{4\pi {\epsilon }_o\left(R+x\right)R} \end{gathered}$$

$$\begin{gathered} \Delta V\left(\frac{R+x}{x}\right)=-\frac{q}{4\pi {\epsilon }_{o}R}\Delta V\left(1+\frac{R}{x}\right) \\ =-\frac{q}{4\pi {\epsilon }_{o}R}\left(1+\frac{R}{x}\right) \\ =-\frac{q}{\left(4\pi {\epsilon }_{o}R\right)\Delta V} \end{gathered}$$

Substitute $0.15\text{m}$ for $R$, $9.0\times {10}^{9}N.m^2/C^2$ for $\frac{1}{4\pi {\epsilon }_o}$, $3.0\times {10}^{-8}\text{C}$ for $q$, and $-500\text{V}$ for $\Delta V$ in the above equation.

$\left(1+\frac{0.15\text{m}}{x}\right)=-\frac{\left(3.0\times {10}^{-8}C\right)\left(9.0\times {10}^{9}N.m^2/C2\right)}{\left(3.0\times {10}^{-8}C\right)\left(-500V\right)}$

$$\begin{gathered} \left(1+\frac{0.15\text{m}}{x}\right)=3.6 \\ \frac{0.15\text{m}}{x}=2.6 \end{gathered}$$

Rewrite the above expression for $x$.

$$\begin{gathered} x=\frac{0.15\text{m}}{2.6} \\ =0.0576\text{m} \\ =5.76\text{cm} \\ \approx 5.8\text{cm} \end{gathered}$$

Hence, the electric potential decreased by $500\text{V}$ at a distance is $5.8\text{cm}$ .