Question

A uniformly charged disk of radius 35.0 cm carries a charge density of

Answer 1

Answer:

a. $E_z = 1.14\times 10^9~{\rm N/C}$

b. $E_z = 9.63\times 10^8~{\rm N/C}$

c. $E_z = 2.4\times 10^8~{\rm N/C}$

d. $E_z = 2.03\times 10^5~{\rm N/C}$

Explanation:

First we calculate the electric field of a uniformly charged disk.

We should choose an infinitesimal area of da on the disk. This area is infinitely small, so it can be regarded as a point. The small portion is chosen arbitrary and a distance 'r' from the center. Then, we will calculate the electric field created by this small area, dE. Finally, we will integrate dE over the disk using cylindrical coordinates.

The electric field formula for point charges is as follows

$d\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{dQ}{r^2}\^r$

Here, dQ can be derived from the charge density of the disk.

$\sigma = \frac{Q}{\pi R^2} = \frac{dQ}{da}\\dQ = \frac{Qda}{\pi R^2}$

Furthermore, dE is a vector, and needs to be separated to its vertical and horizontal components. Otherwise, we cannot integrate.

Let us denote the vertical direction as 'z'. By symmetry, the horizontal components cancel out each other during the integration, such as each component has an equal but opposite counterpart on the other side of the disk.

$dE_z = dE\sin(\theta) = dE\frac{z}{\sqrt{z^2 + r^2}}$

Let's combine above equations:

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{da}{(z^2 + r^2)^{3/2}}$

Now let's integrate over the disk. Since the disk is two dimensional, we will use double integral over the radius, r, and the angle, θ.

$E_z = \int \int dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 {\int\limits^R_0 {\frac{1}{(z^2 + r^2)^{3/2}}} \, rdr } \, d\theta\\E_z = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[1 - \frac{1}{\sqrt{(R^2/z^2) + 1}}]$

Now we can substitute the given variables to calculate the electric field.

(a) z = 0.05 m.

$E_z = \frac{1}{4\pi(8.8\times 10^{-12})}\frac{9\times 10^{-3}}{\pi (0.35)^2}[1 - \frac{1}{\sqrt{((0.35)^2/(0.05)^2) + 1}}] = 1.14\times 10^9~{\rm N/C}$

(b) z = 0.1 m.

$E_z = 9.63\times 10^8~{\rm N/C}$

(c) z = 0.5 m.

$E_z = 2.4\times 10^8~{\rm N/C}$

(d) z = 20 m.

$E_z = 2.03\times 10^5~{\rm N/C}$

Note that the directions of electric fields in each case is in the z-direction (upwards).

Answer 2

Answer:

a.

$E=4.37*10^{8}N/C$

b. $E=3.69*10^{8}N/C$

c. $E_{z} =0.92*10^{8}N/C$

d. $E_{z} =0.76*10^{8}N/C$

Explanation:

To determine the electric field, using the equation describing the electric field

$E=2\pi k\alpha [1-\frac{Z }{(R^{2}+Z^{2} )^{1/2} }]\\K=9*10^{9}NM^{2}/C^{2} \\\alpha =9.0*10^{-3}$

a. For R=35cm = 0.35m and z=5cm =0.05m

we can compute the electric field as

$E=2\pi k\alpha [1-\frac{Z }{(R^{2}+Z^{2})^{\frac{1}{2}} }]\\K=9*10^{9}NM^{2}/C^{2} \\\alpha =9.0*10^{-3}\\\\E=2\pi *9*10^{9}*9.0*10^{-3}[1-\frac{0.05}{(0.35^{2}+0.05^{2})^{\frac{1}{2}}}] \\E=508.9*10^{6}[1-0.14]\\E=508.9*10^{6}*0.86\\E=4.37*10^{8}N/C$

b. For R=35cm = 0.35m and z=10cm =0.1m

$E=2\pi k\alpha [1-\frac{Z }{(R^{2}+Z^{2})^{\frac{1}{2}} }]\\K=9*10^{9}NM^{2}/C^{2} \\\alpha =9.0*10^{-3}\\\\E=2\pi *9*10^{9}*9.0*10^{-3}[1-\frac{0.1}{(0.35^{2}+0.1^{2})^{\frac{1}{2}}}] \\E=508.9*10^{6}[1-0.274]\\E_{z} =508.9*10^{6}*0.73\\E_{z} =3.69*10^{8}N/C$

c. For R=35cm = 0.35m and z=50cm =0.5m

$E=2\pi k\alpha [1-\frac{Z }{(R^{2}+Z^{2})^{\frac{1}{2}} }]\\K=9*10^{9}NM^{2}/C^{2} \\\alpha =9.0*10^{-3}\\\\E=2\pi *9*10^{9}*9.0*10^{-3}[1-\frac{0.5}{(0.35^{2}+0.5^{2})^{\frac{1}{2}}}] \\E=508.9*10^{6}[1-0.819]\\E_{z} =508.9*10^{6}*0.18\\E_{z} =0.92*10^{8}N/C$

d.For R=35cm = 0.35m and z=200cm =2m

$E=2\pi k\alpha [1-\frac{Z }{(R^{2}+Z^{2})^{\frac{1}{2}} }]\\K=9*10^{9}NM^{2}/C^{2} \\\alpha =9.0*10^{-3}\\\\E=2\pi *9*10^{9}*9.0*10^{-3}[1-\frac{2}{(0.35^{2}+2^{2})^{\frac{1}{2}}}] \\E=508.9*10^{6}[1-0.99]\\E_{z} =508.9*10^{6}*0.015\\E_{z} =0.76*10^{8}N/C$