Chapter 25, Problem 3OQ

(i)

To determine

The charge shared between the spheres $\text{A}$ and $\text{B}$.

Answer to Problem 3OQ

Option (c) $150\text{nC}$ on $\text{A}$ and $300\text{nC}$ on $\text{B}$.

Explanation of Solution

The sum of charge on point $\text{A}$ and charge on point $\text{B}$ is equal to $450\text{nC}$.

    $q_{\text{A}}+q_{\text{B}}=450\text{nC}$                                                                                (I)

Write the expression for potential on the sphere $\text{A}$.

    $V_{\text{A}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{A}}}{r_{\text{A}}}$

Here, $V_{\text{A}}$ is the potential on sphere $\text{A}$, $r_{\text{A}}$ is the radius of the sphere $\text{A}$ and $q_{\text{A}}$ is the charge on sphere $\text{A}$.

Write the expression for potential on the sphere $\text{B}$.

    $V_{\text{B}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{B}}}{r_{\text{B}}}$

Here, $V_{\text{B}}$ is the potential on sphere $\text{B}$, $r_{\text{B}}$ is the radius of the sphere $\text{B}$ and $q_{\text{B}}$ is the charge on sphere $\text{B}$.

The potential difference at point $\text{A}$ and $\text{B}$ are equal.

    $V_{\text{A}}=V_{\text{B}}$

Substitute $\frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{A}}}{r_{\text{A}}}$ for $V_{\text{A}}$ and $\frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{B}}}{r_{\text{B}}}$ for $V_{\text{B}}$ in the above equation.

    $\begin{array}{c}\frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{A}}}{r_{\text{A}}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{B}}}{r_{\text{B}}}\\ \frac{q_{\text{A}}}{r_{\text{A}}}=\frac{q_{\text{B}}}{r_{\text{B}}}\\ \frac{q_{\text{A}}}{q_{\text{B}}}=\frac{r_{\text{A}}}{r_{\text{B}}}\end{array}$                                                                            (II)

Conclusion:

Substitute $1.00\text{cm}$ for $r_{\text{A}}$ and $2.00\text{cm}$ for $r_{\text{B}}$ in equation (II).

    $\begin{array}{c}\frac{q_{\text{A}}}{q_{\text{B}}}=\frac{\left(1.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1.00\text{cm}}\right)}{\left(2.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1.00\text{cm}}\right)}\\ \frac{q_{\text{A}}}{q_{\text{B}}}=\frac{1.00\times {10}^{-2}\text{m}}{2.00\times {10}^{-2}\text{m}}\\ \frac{q_{\text{A}}}{q_{\text{B}}}=0.5\\ q_{\text{A}}=0.5q_{\text{B}}\end{array}$

Substitute $0.5q_{\text{B}}$ for $q_{\text{A}}$ in equation (I) to find to find $q_{\text{B}}$.

    $\begin{array}{c}0.5q_{\text{B}}+q_{\text{B}}=450\text{nC}\\ 1.5q_{\text{B}}=450\text{nC}\\ q_{\text{B}}=\frac{450\text{nC}}{1.5}\\ q_{\text{B}}=300\text{nC}\end{array}$

Substitute $300\text{nC}$ for $q_{\text{B}}$ in equation (I) to find $q_{\text{A}}$.

    $\begin{array}{c}{q}_{\text{A}}+300\text{nC}=450\text{nC}\\ q_{\text{A}}=150\text{nC}\end{array}$

The charge at point $\text{A}$ is $150\text{nC}$ and the charge at point $\text{B}$ is $300\text{nC}$.

Therefore, option (c) $150\text{nC}$ on $\text{A}$ and $300\text{nC}$ on $\text{B}$; is correct.

(ii)

To determine

The charge shared between the spheres.

Answer to Problem 3OQ

The charge shared between the spheres is $150\text{nC}$.

Explanation of Solution

Write the expression for potential on the metal sphere $\text{A}$.

    $V_{\text{A}}=\frac{1}{4\pi {\epsilon }_0}\frac{q_{\text{A}}}{r_{\text{A}}}$                                                                                         (III)

The initial potential on the metal sphere $\text{B}$ is zero.

Write the expression for common potential.

    $\Delta V=\frac{Q}{C_{\text{A}}+C_{\text{B}}}$

Here, $\Delta V$ is the common potential, $Q$ is the charge.

Substitute $4\pi {\epsilon }_0{r}_{\text{A}}$ for $C_{\text{A}}$ and $4\pi {\epsilon }_0{r}_{\text{B}}$ for $C_{\text{B}}$ in the above equation.

    $\begin{array}{c}\Delta V=\frac{Q}{4\pi {\epsilon }_0{r}_{\text{A}}+4\pi {\epsilon }_0{r}_{\text{B}}}\\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{Q}{r_{\text{A}}+r_{\text{B}}}\right)\end{array}$                                                                            (IV)

Write the expression for charge on sphere $\text{B}$.

    $q_{\text{B}}=C_{\text{B}}\Delta V$

Substitute $\left(4\pi {\epsilon }_0\right)r_{\text{B}}$ for $C_{\text{B}}$ in the above equation.

    $q_{\text{B}}=\left(4\pi {\epsilon }_0\right)r_{\text{B}}\Delta V$                                                                                  (V)

Write the expression for net charge at point $\text{A}$.

    $q_{\text{net}}=q_{\text{A}}-q_{\text{B}}$                                                                                      (VI)

Conclusion:

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\frac{1}{4\pi {\epsilon }_0}$, $450\text{nC}$ for $q_{\text{A}}$ and $1.00\text{cm}$ for $r_{\text{A}}$ in equation (III) to find $V_{\text{A}}$.

    $\begin{array}{c}{V}_{\text{A}}=\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(450\text{nC}\right)\left(\frac{{10}^{-9}\text{C}}{1.00\text{nC}}\right)}{\left(1.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1.00\text{cm}}\right)}\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\frac{\left(450\times {10}^{-9}\text{C}\right)}{\left(1.00\times {10}^{-2}\text{m}\right)}\\ =4050\times {10}^2\text{V}\end{array}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $\frac{1}{4\pi {\epsilon }_0}$, $450\text{nC}$ for $q_{\text{A}}$, $2.00\text{cm}$ for $q_{\text{B}}$ and $1.00\text{cm}$ for $r_{\text{A}}$ in equation (IV) to find $\Delta V$.

    $\begin{array}{c}\Delta V=\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left[\frac{\left(450\text{nC}\right)\left(\frac{{10}^{-9}\text{C}}{1.00\text{nC}}\right)}{\left(1.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1.00\text{cm}}\right)+\left(2.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1.00\text{cm}}\right)}\right]\\ =\left(9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left[\frac{450\times {10}^{-9}\text{C}}{1.00\times {10}^{-2}\text{m}+2.00\times {10}^{-2}\text{m}}\right]\\ =1350\times {10}^2\text{V}\end{array}$

Substitute $\frac{1}{9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}$ for $4\pi {\epsilon }_0$, $2.00\times {10}^{-2}\text{m}$ for $r_{\text{B}}$ and $1350\times {10}^2\text{V}$ for $\Delta V$ in equation (V) to find $q_{\text{B}}$.

    $\begin{array}{c}{q}_{\text{B}}=\left(\frac{1}{9\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}\right)\left(2.00\times {10}^{-2}\text{m}\right)\left(1350\times {10}^2\text{V}\right)\\ =300\times {10}^{-9}\text{C}\left(\frac{1\text{nC}}{{10}^{-9}\text{C}}\right)\\ =300\text{nC}\end{array}$

Substitute $450\text{nC}$ for $q_{\text{A}}$ and $300\text{nC}$ for $q_{\text{B}}$ in equation (VI) to find $q_{\text{net}}$.

    $\begin{array}{c}{q}_{\text{net}}=450\text{nC}-300\text{nC}\\ =150\text{nC}\end{array}$

Therefore, the charge shared between the spheres is $150\text{nC}$.