Chapter 25, Problem 59AP

(a)

To determine

The radius of the sphere.

Answer to Problem 59AP

The radius of the sphere is $30.0\text{cm}$.

Explanation of Solution

Write the expression of the electric potential on the radius of the sphere.

    $V_{\text{R}}=k_{\text{e}}\left(\frac{q}{R}\right)$                                                                                             (I)

Here, $V_{\text{R}}$ is the electric potential on the radius of the sphere, $k_{\text{e}}$ is the electric constant, $q$ is the charge and $R$ is the radius of the sphere.

Write the expression to obtain the electric potential at the $10.0\text{cm}$ outside the sphere from the centre.

    $\begin{array}{c}{V}_{\text{r}}=k_{\text{e}}\left(\frac{q}{R+10.0\text{cm}}\right)\\ =k_{\text{e}}\left(\frac{q}{R+10.0\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ =k_{\text{e}}\left(\frac{q}{R+0.1\text{m}}\right)\end{array}$                                                                     (II)

Here, $V_{\text{r}}$ is the electric potential at the $10.0\text{cm}$ outside the sphere from the centre.

Divide equation (I) and (II).

    $\begin{array}{c}\frac{V_{\text{R}}}{V_{\text{r}}}=\frac{k_{\text{e}}\left(\frac{q}{R}\right)}{k_{\text{e}}\left(\frac{q}{R+0.1\text{m}}\right)}\\ =\frac{R+0.1\text{m}}{R}\end{array}$

Conclusion:

Substitute $200\text{V}$ for $V_{\text{R}}$ and $150\text{V}$ for $V_{\text{r}}$ in the above equation to calculate $R$.

    $\begin{array}{c}\frac{200\text{V}}{150\text{V}}=\frac{R+0.1\text{m}}{R}\\ 200R=150\left(R+0.1\text{m}\right)\\ 200R=150\left(R+0.1\text{m}\right)\\ =150R+15\text{m}\end{array}$

Further solve the above equation.

    $\begin{array}{c}200R-150R=15\text{m}\\ \text{50}R=15\text{m}\\ R=\frac{15\text{m}}{50}\\ R=0.3\text{m}\end{array}$

Further solve the above equation.

    $\begin{array}{c}R=0.3\text{m}\times \left(\frac{100\text{cm}}{1\text{m}}\right)\\ R=30.0\text{cm}\end{array}$

Therefore, the radius of the sphere is $30.0\text{cm}$.

(b)

To determine

The charge on the sphere.

Answer to Problem 59AP

The charge on the sphere is $6.67\text{nC}$.

Explanation of Solution

Conclusion:

Substitute, $9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2$ for $k_{\text{e}}$, $0.3\text{m}$ for $R$ and $200\text{V}$ for $V_{\text{R}}$ in equation (I) to calculate $q$.

    $\begin{array}{c}200\text{V}=9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2\left(\frac{q}{0.3\text{m}}\right)\\ \left(\frac{q}{0.3\text{m}}\right)=\frac{200\text{V}}{9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2}\\ q=\frac{200\text{V}}{9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2}\times 0.3\text{m}\\ q=6.67\times {10}^{-9}\text{C}\end{array}$

Further solve the above equation.

    $\begin{array}{c}q=6.67\times {10}^{-9}\text{C}\times \frac{{10}^9\text{nC}}{1\text{C}}\\ =6.67\text{nC}\end{array}$

Therefore, the charge on the sphere is $6.67\text{nC}$.

(c)

To determine

The radius of the sphere.

Answer to Problem 59AP

The radius of the sphere can be $29.1\text{cm}\text{or}\text{3}\text{.44}\text{cm}$.

Explanation of Solution

Write the expression to obtain the electric field at the $10.0\text{cm}$ outside the sphere from the centre.

    $\begin{array}{c}{E}_{\text{r}}=k_{\text{e}}\left(\frac{q}{{\left(R+10.0\text{cm}\right)}^2}\right)\\ =k_{\text{e}}\left(\frac{q}{{\left(R+10.0\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\right)\\ =k_{\text{e}}\left(\frac{q}{{\left(R+0.1\text{m}\right)}^2}\right)\end{array}$                                                                   (III)

Here, $E_{\text{r}}$ is the electric field at the $10.0\text{cm}$ outside the sphere from the centre.

Divide equation (I) and (III).

    $\begin{array}{c}\frac{V_{\text{R}}}{E_{\text{r}}}=\frac{k_{\text{e}}\left(\frac{q}{R}\right)}{k_{\text{e}}\left(\frac{q}{{\left(R+0.1\text{m}\right)}^2}\right)}\\ =\frac{{\left(R+0.1\text{m}\right)}^2}{R}\end{array}$

Conclusion:

Substitute $200\text{V}$ for $V_{\text{R}}$ and $400\text{V}/\text{m}$ for $E_{\text{r}}$ in the above equation to calculate $R$.

    $\begin{array}{c}\frac{200\text{V}}{400\text{V}/\text{m}}=\frac{{\left(R+0.1\text{m}\right)}^2}{R}\\ \left(200\text{V}\right)R=\left(400\text{V}/\text{m}\right){\left(R+0.1\text{m}\right)}^2\end{array}$

Further solve the above equation.

    $\begin{array}{c}R=0.291\text{m}\text{or}\text{0}\text{.0344}\text{m}\\ =0.291\text{m}\times \left(\frac{100\text{cm}}{1\text{m}}\right)\text{or}\text{0}\text{.0344}\text{m}\times \left(\frac{100\text{cm}}{1\text{m}}\right)\\ =29.1\text{cm}\text{or}\text{3}\text{.44}\text{cm}\end{array}$

Therefore, the radius of the sphere can be $29.1\text{cm}\text{or}\text{3}\text{.44}\text{cm}$.

(d)

To determine

The charge on the sphere.

Answer to Problem 59AP

The charge on the sphere is $6.79\text{nC}$ when radius of the sphere is $0.291\text{m}$ and the charge on the sphere is $804\text{pC}$ when radius of the sphere is $0.0344\text{m}$.

Explanation of Solution

Conclusion:

Case (i): When radius of the sphere is $0.291\text{m}$.

Substitute, $9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2$ for $k_{\text{e}}$, $0.291\text{m}$ for $R$ and $210\text{V}$ for $V_{\text{R}}$ in equation (I) to calculate $q$.

    $\begin{array}{c}210\text{V}=9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2\left(\frac{q}{0.291\text{m}}\right)\\ \left(\frac{q}{0.291\text{m}}\right)=\frac{210\text{V}}{9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2}\\ q=\frac{210\text{V}}{9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2}\times 0.291\text{m}\\ q=6.79\times {10}^{-9}\text{C}\end{array}$

Further solve the above equation.

    $\begin{array}{c}q=6.79\times {10}^{-9}\text{C}\times \frac{{10}^9\text{nC}}{1\text{C}}\\ =6.79\text{nC}\end{array}$

Case (ii): When radius of the sphere is $0.0344\text{m}$.

Substitute, $9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2$ for $k_{\text{e}}$, $0.0344\text{m}$ for $R$ and $210\text{V}$ for $V_{\text{R}}$ in equation (I) to calculate $q$.

    $\begin{array}{c}210\text{V}=9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2\left(\frac{q}{0.0344\text{m}}\right)\\ \left(\frac{q}{0.0344\text{m}}\right)=\frac{210\text{V}}{9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2}\\ q=\frac{210\text{V}}{9\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^2}\times 0.0344\text{m}\\ q=8.046\times {10}^{-10}\text{C}\end{array}$

Further solve the above equation.

    $\begin{array}{c}q=8.046\times {10}^{-10}\text{C}\times \frac{{10}^{12}\text{pC}}{1\text{C}}\\ =804\text{pC}\end{array}$

Therefore, the charge on the sphere is $6.79\text{nC}$ when radius of the sphere is $0.291\text{m}$ and the charge on the sphere is $804\text{pC}$ when radius of the sphere is $0.0344\text{m}$.

(e)

To determine

Weather the answers in part $\left(c\right)$ and $\left(d\right)$ are unique.

Answer to Problem 59AP

The answers in part $\left(c\right)$ and $\left(d\right)$ are not unique.

Explanation of Solution

The answers in part $\left(c\right)$ are $29.10\text{cm}\text{or}\text{3}\text{.44}\text{cm}$ for radius of the sphere which is not unique.

The answers in part $\left(d\right)$ are $6.79\times {10}^{-9}\text{C}$ or $8.026\times {10}^{-10}\text{C}$ for the charge on the sphere which is also not unique.

Therefore, the answers in part $\left(c\right)$ and $\left(d\right)$ are not unique.