Chapter 20, Problem 31P

(a)

To determine

The electric potential of each sphere.

Answer to Problem 31P

The electric potential of each sphere is $\underset{\_}{1.35\times {10}^5\text{V}}$.

Explanation of Solution

The potential on each sphere will be same.

Write the equation at same potential.

  $V_1=V_2$        (I)

Here, $V_1,V_2$ are the potential for larger and smaller spheres respectively.

Write the expression for the potential larger sphere.

$\left(V_1\right)=\frac{k{q}_1}{r_1}$        (II)

Here, $k$ is the Coulomb’s constant, $q_1$ is the charge of larger sphere and $r_1$ is the radius of larger sphere.

Write the expression for the potential for smaller sphere.

$\left(V_2\right)=\frac{k{q}_2}{r_2}$        (III)

Here, $k$ is the Coulomb’s constant, $q_2$ is the charge of smaller sphere and $r_2$ is the radius of smaller sphere.

Rewrite the expression for the potential from equation (I) by using (II) and (III).

$\begin{array}{l}\frac{k{q}_1}{r_1}=\frac{k{q}_2}{r_2}\\ q_1=\frac{q_2{r}_1}{r_2}\end{array}$        (IV)

Write the expression for total charge by using (IV).

$\begin{array}{l}{q}_{net}=\left(\frac{q_2{r}_1}{r_2}\right)+q_2\\ q_2=\frac{q_{net}}{1+\left(r_1/r_2\right)}\end{array}$        (V)

Conclusion:

Substitute, $1.20\text{μC}$ for $q_{net}$, $6.00\text{cm}$ for $r_1$, $2.00\text{cm}$ for $r_2$ in Equation (V) to find $q_2$.

  $\begin{array}{c}{q}_2=\frac{\left[\left(1.20\text{μC}\right)\left(\frac{1\times {10}^{-6}\text{C}}{1\text{μC}}\right)\right]}{1+\left(\frac{\left[\left(6.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left[\left(2.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right)}\\ =0.300\times {10}^{-6}\text{C}\end{array}$

Substitute, $0.300\times {10}^{-6}\text{C}$ for $q_2$, $6.00\text{cm}$ for $r_1$, $2.00\text{cm}$ for $r_2$ in Equation (IV) to find $q_1$.

$\begin{array}{c}{q}_1=\frac{\left(0.300\times {10}^{-6}\text{C}\right)\left[\left(6.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}{\left[\left(2.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\\ =0.900\times {10}^{-6}\text{C}\end{array}$

Substitute, $0.900\times {10}^{-6}\text{C}$ for $q_1$, $6.00\text{cm}$ for $r_1$, $\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)$ for $k$ in Equation (II) to find $V_1$.

$\begin{array}{c}{V}_1=\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(0.900\times {10}^{-6}\text{C}\right)}{\left[\left(6.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\\ =1.35\times {10}^5\text{V}\end{array}$

The sphere is having same potential.

Thus, the electric potential of each sphere is $\underset{\_}{1.35\times {10}^5\text{V}}$.

(b)

To determine

The electric field at the surface of each sphere.

Answer to Problem 31P

The electric fields at the surface of each sphere are $2.25\times {10}^6\text{V/m , 6}\text{.74}\times {\text{10}}^6\text{V/m}$.

Explanation of Solution

Write the equation electric field for larger sphere.

  $E_1=\frac{V_1}{r_1}$        (VI)

Here, $E_1$ is the electric field for larger sphere.

Write the equation electric field for smaller sphere.

  $E_2=\frac{V_2}{r_2}$        (VII)

Here, $E_2$ is the electric field for smaller sphere.

Conclusion:

Substitute, $1.35\times {10}^5\text{V}$ for $V_1$, $6.00\text{cm}$ for $r_1$, in Equation (VI) to find $E_1$.

  $\begin{array}{c}{E}_1=\frac{\left(1.35\times {10}^5\text{V}\right)}{\left[\left(6.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\\ =2.25\times {10}^6\text{V/m}\end{array}$

Substitute, $1.35\times {10}^5\text{V}$ for $V_2$, $2.00\text{cm}$ for $r_2$, in Equation (VII) to find $E_2$.

  $\begin{array}{c}{E}_2=\frac{\left(1.35\times {10}^5\text{V}\right)}{\left[\left(2.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\\ =6.74\times {10}^6\text{V/m}\end{array}$

Thus, the electric fields at the surface of each sphere are $2.25\times {10}^6\text{V/m , 6}\text{.74}\times {\text{10}}^6\text{V/m}$.