Chapter 24, Problem 55AP

To determine

The graph of the magnitude of electric field due to the given configuration versus $r$ for $0<r<25.0\text{cm}$.

Answer to Problem 55AP

The graph of the magnitude of electric field due to the given configuration versus $r$ for $0<r<25.0\text{cm}$ is shown in figure below.

Explanation of Solution

Write the expression to calculate the electric field on uniformly charged sphere for $r<R$.

    $E=k_{\text{e}}\frac{rQ}{R^3}$                       (I)

Here, $E$ is the electric field, $Q$ is the charge, $k_{\text{e}}$ is the Coulomb’s constant $R$ is the radius of insulating sphere and $r$ is the distance between the point.

Write the expression to calculate the electric field on uniformly charged sphere for $r\ge R$.

    $E=k_{\text{e}}\frac{Q}{r^2}$                       (II)

The electric field inside a conductor is $0$.

So, for the region $10.00\text{cm}<r<15.00\text{cm}$.

    $E=0$.

Write the expression to calculate the net charge outside the sphere.

    $Q_{\text{net}}=Q_{\text{out}}-Q$                     (III)

Here, $Q_{\text{net}}$ is the net charge, $Q$ is the charge inside the sphere, $Q_{\text{out}}$ is the charge outside the sphere.

 Write the expression to calculate the electric field outside the uniformly charged sphere.

    $E=k_{\text{e}}\frac{Q_{\text{out}}}{r^2}$                    (IV)

Conclusion:

For $r\le 5.00\text{cm}$

Assume $r=2.50\text{cm}$.

Substitute $9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $2.50\text{cm}$ for $r$, $3.00\text{μC}$ for $Q$ and $5.00\text{cm}$ for $R$ in equation (I) to solve for $E$.

    $\begin{array}{c}E=9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}\frac{\left(2.50\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(3.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(5.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^3}\\ =5.4\times {10}^6\text{N}/\text{C}\times \frac{{10}^{-6}\text{MN}}{1\text{N}}\\ =5.4\text{MN}/\text{C}\end{array}$

Calculate the electric field at $r=5.00\text{cm}$.

Substitute $9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $5.00\text{cm}$ for $r$, $3.00\text{μC}$ for $Q$ and $5.00\text{cm}$ for $R$ in equation (II) to solve for $E$.

    $\begin{array}{c}E=9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}\frac{\left(5.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(3.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(5.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^3}\\ =10.8\times {10}^6\text{N}/\text{C}\times \frac{{10}^{-6}\text{MN}}{1\text{N}}\\ =10.8\text{MN}/\text{C}\end{array}$

For region $5.00\text{cm}<r<\text{10}\text{.00}\text{cm}$.

At $r=7.50\text{cm}$

Substitute $9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $7.50\text{cm}$ for $r$, $3.00\text{μC}$ for $Q$ in equation (II) to solve for $E$.

    $\begin{array}{c}E=9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}\frac{\left(3.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(7.50\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2}\\ =4.8\times {10}^6\text{N}/\text{C}\times \frac{{10}^{-6}\text{MN}}{1\text{N}}\\ =4.8\text{MN}/\text{C}\end{array}$

At $r=10.00\text{cm}$,

Substitute $9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $10.00\text{cm}$ for $r$, $3.00\text{μC}$ for $Q$ in equation (II) to solve for $E$.

    $\begin{array}{c}E=9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}\frac{\left(3.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(10.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2}\\ =2.7\times {10}^6\text{N}/\text{C}\times \frac{{10}^{-6}\text{MN}}{1\text{N}}\\ =2.7\text{MN}/\text{C}\end{array}$

For region $r>\text{15}\text{.00}\text{cm}$,

Substitute $-1.00\text{μC}$ for $Q_{\text{net}}$, $3.00\text{μC}$ for $Q$ in equation (III) to solve for $Q_{\text{out}}$.

    $\begin{array}{l}-1.00\text{μC}=Q_{\text{out}}-\\ Q_{\text{out}}=2.00\text{μC}\end{array}$

At $r=25.00\text{cm}$,

    $\begin{array}{c}E=9\times {10}^9{\text{N-m}}^{\text{2}}/{\text{C}}^{\text{2}}\frac{\left(2.00\text{μC}\times \frac{{10}^{-6}\text{C}}{1\text{μC}}\right)}{{\left(25.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2}\\ =0.288\times {10}^6\text{N}/\text{C}\times \frac{{10}^{-6}\text{MN}}{1\text{N}}\\ =0.288\text{MN}/\text{C}\end{array}$

The graph of the magnitude of electric field due to the given configuration versus $r$ for $0<r<25.0\text{cm}$ is shown in figure below.

Figure (1)