Question

a ring of radius 2.50 cm has a uniform linear charge density of 15.3 nc/m. what are the magnitude and direction of the electric field produced by the ring at a point at its central axis at a distance of 11.9 cm from the center of the ring?

o 1.43 × 10³ n/c, along the axis and away from the center of the ring
o 2.86 × 10³ n/c, along the axis and toward the center of the ring
o 1.43 × 10³ n/c, along the axis and toward the center of the ring
o 1.14 × 10³ n/c, along the axis and toward the center of the ring
o 2.86 × 10³ n/c, along the axis and away from the center of the ring

Explanation:

Step1: Calculate the total charge $Q$

The circumference of the ring is $C = 2\pi r$, and the total charge $Q=\lambda C$, where $\lambda = 15.3\times10^{- 9}\text{ C/m}$ and $r = 2.50\times10^{-2}\text{ m}$. So $C = 2\pi\times2.50\times10^{-2}\text{ m}$, and $Q=\lambda\times2\pi r=(15.3\times10^{-9}\text{ C/m})\times2\pi\times2.50\times10^{-2}\text{ m}$.

Step2: Use the formula for electric - field on the axis of a charged ring

The formula for the electric field on the axis of a charged ring at a distance $x$ from the center of the ring is $E=\frac{kQx}{(x^{2}+r^{2})^{\frac{3}{2}}}$, where $k = 9\times10^{9}\text{ N}\cdot\text{m}^{2}/\text{C}^{2}$, $Q$ is the total charge of the ring, $x = 11.9\times10^{-2}\text{ m}$, and $r = 2.50\times10^{-2}\text{ m}$.
First, calculate $Q=(15.3\times10^{-9}\text{ C/m})\times2\pi\times2.50\times10^{-2}\text{ m}\approx2.40\times10^{-9}\text{ C}$.
Then, substitute the values into the electric - field formula:
\[

$$\begin{align*} E&=\frac{(9\times10^{9}\text{ N}\cdot\text{m}^{2}/\text{C}^{2})\times(2.40\times10^{-9}\text{ C})\times(11.9\times10^{-2}\text{ m})}{((11.9\times10^{-2}\text{ m})^{2}+(2.50\times10^{-2}\text{ m})^{2})^{\frac{3}{2}}}\\ &=\frac{(9\times10^{9})\times(2.40\times10^{-9})\times(11.9\times10^{-2})}{((11.9\times10^{-2})^{2}+(2.50\times10^{-2})^{2})^{\frac{3}{2}}}\text{ N/C}\\ \end{align*}$$

\]
\[

$$\begin{align*} (11.9\times10^{-2})^{2}+(2.50\times10^{-2})^{2}&=1.4161\times10^{-2}+6.25\times10^{-4}\\ &=1.4161\times10^{-2}+0.0625\times10^{-2}\\ &=1.4786\times10^{-2}\text{ m}^{2} \end{align*}$$

\]
\[

$$\begin{align*} (1.4786\times10^{-2})^{\frac{3}{2}}&=(1.4786)^{\frac{3}{2}}\times10^{-3}\\ &\approx1.77\times10^{-3}\text{ m}^{3} \end{align*}$$

\]
\[

$$\begin{align*} E&=\frac{(9\times10^{9})\times(2.40\times10^{-9})\times(11.9\times10^{-2})}{1.77\times10^{-3}}\text{ N/C}\\ &=\frac{9\times2.40\times11.9\times10^{-2}}{1.77}\text{ N/C}\\ &=\frac{257.04\times10^{-2}}{1.77}\text{ N/C}\\ &\approx1.43\times10^{3}\text{ N/C} \end{align*}$$

\]
Since the charge is positive, the direction of the electric field is along the axis and away from the center of the ring.

Answer:

A. $1.43\times 10^{3}\text{ N/C}$, along the axis and away from the center of the ring