Chapter 19, Problem 15P

(a)

To determine

The electric field on the axis of a ring at $1.00\text{cm}$.

Answer to Problem 15P

The electric field on the axis of a ring at $1.00\text{cm}$ is $6.64\times {10}^6\text{N}/\text{C}$ away from the center of the ring.

Explanation of Solution

Given info: The radius of the uniformly charged ring is $10.0\text{cm}$ and the total charge on the ring is $75.0\mu \text{C}$.

From the Coulomb’s law the formula to calculate the electric field at any distance on the axis due to a uniformly charged ring.

${\overrightarrow{E}}_x=\frac{Kqx}{{\left(r^2+x^2\right)}^{\frac{3}{2}}}$ (1)

Here,

$K$ is the proportionality constant.

$r$ is the radius of the ring.

$x$ is the distance of the electric field point from the center of the ring.

The proportionality constant $K$ is,

$K=\frac{1}{4\pi {\epsilon }_0}$

Here,

${\epsilon }_0$ is the absolute electrical permeability.

Substitute $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$ in the above equation.

$\begin{array}{c}K=\frac{1}{4\pi \left(8.85\times {10}^{-12}\text{F}/\text{m}\right)}\\ =8.991\times {10}^9\text{N}\end{array}$

The charge on the ring is positive so the direction of the electric field is away from the center of the ring.

Substitute $1.00\text{cm}$ for $x$, $10.0\text{cm}$ for $r$, $8.991\times {10}^9\text{N}$ for $K$ and $75.0\mu \text{C}$ for $q$ in the above equation.

$\begin{array}{c}{\overrightarrow{E}}_x=\frac{\left(8.991\times {10}^9\text{N}\right)\left(75.0\mu \text{C}\right)\left(1.00\text{cm}\right)}{{\left({\left(10.0\text{cm}\right)}^2+{\left(1.00\text{cm}\right)}^2\right)}^{\frac{3}{2}}}\\ =\frac{\left(8.991\times {10}^9\text{N}\right)\left(75.0\mu \text{C}\right)\left(\frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)\left(1.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{{\left(101{\text{cm}}^2\right)}^{\frac{3}{2}}{\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^3}\\ =0.6643\times {10}^7\text{N}/\text{C}\\ \approx 6.64\times {10}^6\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the electric field on the axis of a ring at $1.00\text{cm}$ is $6.64\times {10}^6\text{N}/\text{C}$ away from the center of the ring.

(b)

To determine

The electric field on the axis of a ring at $5.00\text{cm}$.

Answer to Problem 15P

The electric field on the axis of a ring at $5.00\text{cm}$ is $2.41\times {10}^7\text{N}/\text{C}$ away from the center of the ring.

Explanation of Solution

Given info: The radius of the uniformly charged ring is $10.0\text{cm}$ and the total charge on the ring is $75.0\mu \text{C}$.

The charge on the ring is positive so the direction of the electric field is away from the center of the ring.

Substitute $5.00\text{cm}$ for $x$, $10.0\text{cm}$ for $r$, $8.991\times {10}^9\text{N}$ for $K$ and $75.0\mu \text{C}$ for $q$ in equation (1).

$\begin{array}{c}{\overrightarrow{E}}_x=\frac{\left(8.991\times {10}^9\text{N}\right)\left(75.0\mu \text{C}\right)\left(5.00\text{cm}\right)}{{\left({\left(10.0\text{cm}\right)}^2+{\left(5.00\text{cm}\right)}^2\right)}^{\frac{3}{2}}}\\ =\frac{\left(8.991\times {10}^9\text{N}\right)\left(75.0\mu \text{C}\right)\left(\frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)\left(5.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{{\left(125{\text{cm}}^2\right)}^{\frac{3}{2}}{\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^3}\\ =2.4125\times {10}^7\text{N}/\text{C}\\ \approx 2.41\times {10}^7\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the electric field on the axis of a ring at $5.00\text{cm}$ is $2.41\times {10}^7\text{N}/\text{C}$ away from the center.

(c)

To determine

The electric field on the axis of a ring at $30.0\text{cm}$.

Answer to Problem 15P

The electric field on the axis of a ring at $30.0\text{cm}$ is $6.39\times {10}^6\text{N}/\text{C}$ away from the center of the ring.

Explanation of Solution

Given info: The radius of the uniformly charged ring is $10.0\text{cm}$ and the total charge on the ring is $75.0\mu \text{C}$.

The charge on the ring is positive so the direction of the electric field is away from the center of the ring.

Substitute $30.0\text{cm}$ for $x$, $10.0\text{cm}$ for $r$, $8.991\times {10}^9\text{N}$ for $K$ and $75.0\mu \text{C}$ for $q$ in the above equation.

$\begin{array}{c}{\overrightarrow{E}}_x=\frac{\left(8.991\times {10}^9\text{N}\right)\left(75.0\mu \text{C}\right)\left(30.0\text{cm}\right)}{{\left({\left(10.0\text{cm}\right)}^2+{\left(30.0\text{cm}\right)}^2\right)}^{\frac{3}{2}}}\\ =\frac{\left(8.991\times {10}^9\text{N}\right)\left(75.0\mu \text{C}\right)\left(\frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)\left(30.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{{\left(1000{\text{cm}}^2\right)}^{\frac{3}{2}}{\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^3}\\ =0.6397\times {10}^7\text{N}/\text{C}\\ \approx 6.39\times {10}^6\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the electric field on the axis of a ring at $30.0\text{cm}$ is $6.39\times {10}^6\text{N}/\text{C}$ away from the center.

(d)

To determine

The electric field on the axis of a ring at $100\text{cm}$.

Answer to Problem 15P

The electric field on the axis of a ring at $100\text{cm}$ is $6.64\times {10}^5\text{N}/\text{C}$ away from the center of the ring.

Explanation of Solution

Given info: The radius of the uniformly charged ring is $10.0\text{cm}$ and the total charge on the ring is $75.0\mu \text{C}$.

The charge on the ring is positive so the direction of the electric field is away from the center of the ring.

Substitute $100\text{cm}$ for $x$, $10.0\text{cm}$ for $r$, $8.991\times {10}^9\text{N}$ for $K$ and $75.0\mu \text{C}$ for $q$ in the above equation.

$\begin{array}{c}{\overrightarrow{E}}_x=\frac{\left(8.991\times {10}^9\text{N}\right)\left(75.0\mu \text{C}\right)\left(100\text{cm}\right)}{{\left({\left(10.0\text{cm}\right)}^2+{\left(100\text{cm}\right)}^2\right)}^{\frac{3}{2}}}\\ =\frac{\left(8.991\times {10}^9\text{N}\right)\left(75.0\mu \text{C}\right)\left(\frac{{10}^{-6}\text{C}}{1\mu \text{C}}\right)\left(100\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{{\left(10100{\text{cm}}^2\right)}^{\frac{3}{2}}{\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^3}\\ =0.0664\times {10}^7\text{N}/\text{C}\\ =6.64\times {10}^5\text{N}/\text{C}\end{array}$

Conclusion:

Therefore, the electric field on the axis of a ring at $100\text{cm}$ is $6.64\times {10}^5\text{N}/\text{C}$ away from the center.