Chapter 24, Problem 57AP

(a)

To determine

The charge on the insulating sphere.

Answer to Problem 57AP

The charge on the insulating sphere is $-4.01\text{nC}$.

Explanation of Solution

Draw a Gaussian sphere from the centre of the insulating sphere such that the radius of the Gaussian sphere is between $a$ and $b$.

Figure-(1)

Here, $r$ is the radius of the Gaussian sphere.

Write the expression to calculate the electric field at the radial distance $r$.

    $\begin{array}{l}E=\frac{k_{\text{c}}\times q_{\text{en}}}{r^2}\\ q_{\text{en}}=\frac{E\times r^2}{k_{\text{c}}}\end{array}$

Here, $E$ is the electric field, $k_{\text{c}}$ is the Coulomb’s constant and $q_{\text{en}}$ is the enclosed charge in the Gaussian sphere.

Conclusion:

Convert the units of $r$ from $\text{cm}$ to $\text{m}$.

    $\begin{array}{c}r=\left(10.0\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.10\text{m}\end{array}$

Substitute $3.60\times {10}^3\text{N}/\text{C}$ for $E$, $0.10\text{m}$ for $r$ and $8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{c}}$ in the above equation to calculate $q_{\text{en}}$.

    $\begin{array}{c}{q}_{\text{en}}=\frac{\left(3.60\times {10}^3\text{N}/\text{C}\right){\left(0.10\text{m}\right)}^2}{8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}\\ =\frac{36\text{N}\cdot {\text{m}}^2/\text{C}}{8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}\\ =\left(4.01\times {10}^{-9}\text{C}\right)\left(\frac{{10}^9\text{nC}}{1\text{C}}\right)\\ =4.01\text{nC}\end{array}$

The electric field at $10.0\text{cm}$ from the centre is radially inwards, thus, the charge on the insulating sphere is negative.

Therefore, the charge on the insulating sphere is $-4.01\text{nC}$.

(b)

To determine

The net charge on the hollow conducting sphere.

Answer to Problem 57AP

The net charge on the hollow conducting sphere is $+9.57\text{nC}$.

Explanation of Solution

Draw a Gaussian sphere from the centre of the insulating sphere such that the radius of the Gaussian sphere is greater than $b$.

Figure-(2)

Here, $r$ is the radius of the Gaussian sphere.

Write the expression to calculate the electric field at the radial distance $r$.

    $\begin{array}{l}E=\frac{k_{\text{c}}\times q_{\text{en}}^{\text{'}}}{r^2}\\ q_{\text{en}}^{\text{'}}=\frac{E\times r^2}{k_{\text{c}}}\end{array}$                                                                                                                (I)

Here, $E$ is the electric field, $k_{\text{c}}$ is the Coulomb’s constant and $q_{\text{en}}^{\text{'}}$ is the enclosed charge in the Gaussian sphere.

Write the expression to calculate the total charge.

    $\begin{array}{l}{q}_{\text{en}}^{\text{'}}=q_{\text{i}}+q_{\text{c}}\\ q_{\text{c}}=q_{\text{en}}^{\text{'}}-q_{\text{i}}\end{array}$                                                                                                              (II)

Here, $q_{\text{en}}^{\text{'}}$ is the total charge, $q_{\text{i}}$ is the charge on the insulating sphere and $q_{\text{c}}$ is the charge on the hollow conducting sphere.

Conclusion:

Convert the units of $r$ from $\text{cm}$ to $\text{m}$.

    $\begin{array}{c}r=\left(50.0\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.50\text{m}\end{array}$

Substitute $200\text{N}/\text{C}$ for $E$, $0.50\text{m}$ for $r$ and $8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{c}}$ in equation (I) to calculate $q_{\text{en}}^{\text{'}}$.

    $\begin{array}{c}{q}_{\text{en}}^{\text{'}}=\frac{\left(200\text{N}/\text{C}\right){\left(0.50\text{m}\right)}^2}{8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}\\ =\frac{50\text{N}\cdot {\text{m}}^2/\text{C}}{8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2}\\ =\left(5.56\times {10}^{-9}\text{C}\right)\left(\frac{{10}^9\text{nC}}{1\text{C}}\right)\\ =5.56\text{nC}\end{array}$

The electric field at $50.0\text{cm}$ from the centre is radially outwards, thus, the total charge is positive.

Substitute $5.56\text{nC}$ for $q_{\text{en}}^{\text{'}}$ and $-4.01\text{nC}$ for $q_{\text{i}}$ in equation (II) to calculate the value of $q_{\text{c}}$.

    $\begin{array}{c}{q}_{\text{c}}=5.56\text{nC}-\left(-4.01\text{nC}\right)\\ =5.56\text{nC}+4.01\text{nC}\\ =9.57\text{nC}\end{array}$

Therefore, the net charge on the hollow conducting sphere is $+9.57\text{nC}$.

(c)

To determine

The charge on the inner surface of the hollow conducting sphere.

Answer to Problem 57AP

The charge on the inner surface of the hollow conducting sphere is $+4.01\text{nC}$.

Explanation of Solution

The insulating sphere induces a charge on the inner surface of the hollow conducting sphere that has the same magnitude but has an opposite sign.

Therefore, the charge on the inner surface of the hollow conducting sphere is $+4.01\text{nC}$.

(d)

To determine

The charge on the outer surface of the hollow conducting sphere.

Answer to Problem 57AP

The charge on the outer surface of the hollow conducting sphere is $+5.56\text{nC}$.

Explanation of Solution

The charge on the outer surface of the hollow conducting sphere is equal to the net charge on the hollow conducting sphere minus the charge on the inner surface of the hollow conducting sphere.

Write the expression to calculate the charge on the outer surface of the hollow conducting sphere.

    $q_{\text{oc}}=q_{\text{c}}-q_{\text{ic}}$

Here, $q_{\text{oc}}$ is the charge on the outer surface of the hollow conducting sphere and $q_{\text{ic}}$ is the charge on the inner surface of the hollow conducting sphere.

Conclusion:

Substitute $9.57\text{nC}$ for $q_{\text{c}}$ and $4.01\text{nC}$ for $q_{\text{ic}}$ in the above equation to calculate the value of $q_{\text{oc}}$.

    $\begin{array}{c}{q}_{\text{oc}}=9.57\text{nC}-4.01\text{nC}\\ =5.56\text{nC}\end{array}$

Therefore, the charge on the outer surface of the hollow conducting sphere is $+5.56\text{nC}$.