Chapter 23, Problem 3P

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 μC. Find the electric field on the axis of the ring at (a) 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, and (d) 100 cm from the center of the ring.

To determine

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

Answer to Problem 3P

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.

To determine

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

Answer to Problem 3P

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.

To determine

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

Answer to Problem 3P

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.

To determine

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

Answer to Problem 3P

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.