Chapter 24, Problem 68PQ

(a)

To determine

The magnitude of electric field along the disk’s axis at a distance of $2.50\text{cm}$.

Answer to Problem 68PQ

The electric field at distance $2.50\text{cm}$ is $1.25\times {10}^9\text{N/C}$.

Explanation of Solution

Write the formula for the electric field created by a disk along the axis of the disk.

    $E=2\pi k\sigma \left(1-\frac{y}{\sqrt{R^2+y^2}}\right)$

Here, $E$ is the electric field, $k$ is the coulomb’s constant, $\sigma$ is the surface charge density, $y$ is the distance of the point from the center, and $R$ is the radius of the disk.

Conclusion:

  Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}$ for $\sigma$, $0.250\text{m}$ for $R$, $2.50\text{cm}$ for $y$ to

determine the electric field.

    $\begin{array}{c}E=2\pi \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}\right)\left(1-\frac{2.50\text{cm}}{\sqrt{{\left(0.250\text{m}\right)}^2+{\left(2.50\text{cm}\right)}^2}}\right)\\ =2\pi \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}\right)\left(1-\frac{2.50\times {10}^{-2}\text{m}}{\sqrt{{\left(0.250\text{m}\right)}^2+{\left(2.50\times {10}^{-2}\text{m}\right)}^2}}\right)\\ =1.25\times {10}^9\text{N/C}\end{array}$

The electric field at distance $2.50\text{cm}$ is $1.25\times {10}^9\text{N/C}$.

(b)

To determine

The magnitude of electric field along the disk’s axis at a distance of $25.0\text{cm}$.

Answer to Problem 68PQ

The electric field at distance $25.0\text{cm}$ is $4.05\times {10}^8\text{N/C}$.

Explanation of Solution

Write the formula for the electric field created by a disk along the axis of the disk.

    $E=2\pi k\sigma \left(1-\frac{y}{\sqrt{R^2+y^2}}\right)$

Here, $E$ is the electric field, $k$ is the coulomb’s constant, $\sigma$ is the surface charge density, $y$ is the distance of the point from the center, and $R$ is the radius of the disk.

Conclusion:

  Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}$ for $\sigma$, $0.250\text{m}$ for $R$, $25.0\text{cm}$ for $y$ to

determine the electric field.

    $\begin{array}{c}E=2\pi \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}\right)\left(1-\frac{25.0\text{cm}}{\sqrt{{\left(0.250\text{m}\right)}^2+{\left(2.50\text{cm}\right)}^2}}\right)\\ =2\pi \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}\right)\left(1-\frac{25.0\times {10}^{-2}\text{m}}{\sqrt{{\left(0.250\text{m}\right)}^2+{\left(25.0\times {10}^{-2}\text{m}\right)}^2}}\right)\\ =4.05\times {10}^8\text{N/C}\end{array}$

The electric field at distance $25.0\text{cm}$ is $4.05\times {10}^8\text{N/C}$.

(c)

To determine

The magnitude of electric field along the disk’s axis at a distance of $50.0\text{cm}$.

Answer to Problem 68PQ

The electric field at distance $50.0\text{cm}$ is $1.46\times {10}^8\text{N/C}$.

Explanation of Solution

Write the formula for the electric field created by a disk along the axis of the disk.

    $E=2\pi k\sigma \left(1-\frac{y}{\sqrt{R^2+y^2}}\right)$

Here, $E$ is the electric field, $k$ is the coulomb’s constant, $\sigma$ is the surface charge density, $y$ is the distance of the point from the center, and $R$ is the radius of the disk.

Conclusion:

  Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}$ for $\sigma$, $0.250\text{m}$ for $R$, $50.0\text{cm}$ for $y$ to

determine the electric field.

  $\begin{array}{c}E=2\pi \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}\right)\left(1-\frac{50.0\text{cm}}{\sqrt{{\left(0.250\text{m}\right)}^2+{\left(50.0\text{cm}\right)}^2}}\right)\\ =2\pi \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}\right)\left(1-\frac{50.0\times {10}^{-2}\text{m}}{\sqrt{{\left(0.250\text{m}\right)}^2+{\left(50.0\times {10}^{-2}\text{m}\right)}^2}}\right)\\ =1.46\times {10}^8\text{N/C}\end{array}$

The electric field at distance $50.0\text{cm}$ is $1.46\times {10}^8\text{N/C}$.

(d)

To determine

The magnitude of electric field along the disk’s axis at a distance of $5.00\text{m}$.

Answer to Problem 68PQ

The electric field at distance $5.00\text{m}$ is $1.73\times {10}^6\text{N/C}$.

Explanation of Solution

Write the formula for the electric field created by a disk along the axis of the disk.

    $E=2\pi k\sigma \left(1-\frac{y}{\sqrt{R^2+y^2}}\right)$

Here, $E$ is the electric field, $k$ is the coulomb’s constant, $\sigma$ is the surface charge density, $y$ is the distance of the point from the center, and $R$ is the radius of the disk.

Conclusion:

  Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}$ for $\sigma$, $0.250\text{m}$ for $R$, $5.00\text{m}$ for $y$ to

determine the electric field.

    $\begin{array}{c}E=2\pi \left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.45\times {10}^{-2}{\text{C/m}}^{\text{2}}\right)\left(1-\frac{5.00\text{m}}{\sqrt{{\left(0.250\text{m}\right)}^2+{\left(5.00\text{m}\right)}^2}}\right)\\ =1.73\times {10}^6\text{N/C}\end{array}$

The electric field at distance $5.00\text{m}$ is $1.73\times {10}^6\text{N/C}$.