Question

a 9.94 -cm -diameter plastic disk is charged uniformly with an extra charge of 16.0 nc. the electric field at a point on the central axis has magnitude of 1.04 × 10^5 n/c. how far is this point from the center of the disk?

o 7.49 mm
o 4.28 mm
o 6.42 mm
o 3.21 mm
o 5.35 mm

Explanation:

Step1: Calculate the radius of the disk

The diameter $d = 9.94\ cm=0.0994\ m$, so the radius $R=\frac{d}{2}=\frac{0.0994}{2}= 0.0497\ m$. The charge $Q = 16.0\ nC=16.0\times10^{-9}\ C$, and the electric - field magnitude $E = 1.04\times10^{5}\ N/C$. The formula for the electric field on the central axis of a charged disk is $E = 2\pi k\sigma(1-\frac{z}{\sqrt{z^{2}+R^{2}}})$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, and $\sigma=\frac{Q}{\pi R^{2}}$.
First, calculate the surface - charge density $\sigma=\frac{Q}{\pi R^{2}}=\frac{16.0\times 10^{-9}}{\pi\times(0.0497)^{2}}$.

Step2: Rearrange the electric - field formula to solve for $z$

The electric - field formula $E = 2\pi k\sigma(1-\frac{z}{\sqrt{z^{2}+R^{2}}})$ can be rewritten as $\frac{E}{2\pi k\sigma}=1 - \frac{z}{\sqrt{z^{2}+R^{2}}}$, then $\frac{z}{\sqrt{z^{2}+R^{2}}}=1-\frac{E}{2\pi k\sigma}$.
Square both sides: $\frac{z^{2}}{z^{2}+R^{2}}=(1 - \frac{E}{2\pi k\sigma})^{2}$.
Cross - multiply: $z^{2}=(z^{2}+R^{2})(1 - \frac{E}{2\pi k\sigma})^{2}$.
Expand the right - hand side: $z^{2}=z^{2}(1 - \frac{E}{2\pi k\sigma})^{2}+R^{2}(1 - \frac{E}{2\pi k\sigma})^{2}$.
Rearrange to get $z^{2}[1-(1 - \frac{E}{2\pi k\sigma})^{2}]=R^{2}(1 - \frac{E}{2\pi k\sigma})^{2}$.
Then $z^{2}=\frac{R^{2}(1 - \frac{E}{2\pi k\sigma})^{2}}{1-(1 - \frac{E}{2\pi k\sigma})^{2}}$, and $z=\sqrt{\frac{R^{2}(1 - \frac{E}{2\pi k\sigma})^{2}}{1-(1 - \frac{E}{2\pi k\sigma})^{2}}}$.
Substitute $\sigma=\frac{Q}{\pi R^{2}}$ into the formula:
\[

$$\begin{align*} 2\pi k\sigma&=2\pi\times9\times 10^{9}\times\frac{16.0\times 10^{-9}}{\pi\times(0.0497)^{2}}\\ &=\frac{2\times9\times16.0}{(0.0497)^{2}}\\ \end{align*}$$

\]
\[

$$\begin{align*} 1-\frac{E}{2\pi k\sigma}&=1-\frac{1.04\times 10^{5}\times(0.0497)^{2}}{2\times9\times16.0}\\ \end{align*}$$

\]
After calculating, we find $z = 6.42\times10^{-3}\ m = 6.42\ mm$.

Answer:

$6.42\ mm$