Question

A uniformly charged disk with radius R = 40.0 cm and uniform charge density σ = 7.20 ✕ 10−3 C/m2 lies in the xy-plane, with its center at the origin. What is the electric field (in MN/C) due to the charged disk at the following locations? (a) z = 5.00 cm MN/C (b) z = 10.0 cm MN/C (c) z = 50.0 cm MN/C (d) z = 200 cm MN/C

Answer 1

Answer:

(a) 356.313 MN/C

(b) 307.909 MN/C

(c) 89.078 MN/C

(d) 7.728 MN/C

Explanation:

$E_{z} = K{\sigma}2\pi(1-\frac{Z}{\sqrt{Z^{2} +R^{2}}})$

where;

$E_{z}$ is Electric field due to location "Z"

K is coloumb's constant = 8.99 X 10⁹ Nm²/C²

${\sigma}$ is charge density, = 7.20 X 10⁻³ C/m²

R is the radius of the charged disk = 40.0cm

Step 1: calculate the electric field due to location 5.0 cm

$E_{z} = K{\sigma}2\pi(1-\frac{Z}{\sqrt{Z^{2} +R^{2}}})$

$E_{5} = 8.99 X 10^9 X 7.20 X 10^{-3} X 2\pi(1-\frac{5}{\sqrt{5^{2}+40^{2}}})$

$E_{5} =4.0675 X 10^8(1-\frac{5}{\sqrt{5^{2} +40^{2}}})$

$E_{5} =4.0675 X 10^8(1- 0.124)$

$E_{5} =4.0675 X 10^8(0.876)$

$E_{5} = 3.56313 X10^8 \frac{N}{C}$ = 356.313 MN/C

Step 2: calculate the electric field due to location 10.0 cm

$E_{10} =4.0675 X 10^8(1-\frac{10}{\sqrt{10^{2} +40^{2}}})$

$E_{10} =4.0675 X 10^8(1- 0.243)$

$E_{10} =4.0675 X 10^8(0.757)$

$E_{10} = 3.07909 X10^8 \frac{N}{C}$ = 307.909 MN/C

Step 3: calculate the electric field due to location 50.0 cm

$E_{50} =4.0675 X 10^8(1-\frac{50}{\sqrt{50^{2}+40^{2}}})$

$E_{50} =4.0675 X 10^8(1- 0.781)$

$E_{50} =4.0675 X 10^8(0.219)$

$E_{50} = 8.90783 X10^7 \frac{N}{C}$ = 89.078 MN/C

Step 4: calculate the electric field due to location 200.0 cm

$E_{200} =4.0675 X 10^8(1-\frac{200}{\sqrt{200^{2} +40^{2}}})$

$E_{200} =4.0675 X 10^8(1- 0.981)$

$E_{200} =4.0675 X 10^8(0.019)$

$E_{200} = 7.72825 X10^6 \frac{N}{C}$ = 7.728 MN/C

Answer 2

(a) The elctric field at z = 5cm is 356 MN/C

(b) The elctric field at z = 10cm is 308 MN/C

(c) The elctric field at z = 50cm is 89 MN/C

(d) The elctric field at z = 200cm is 7.7 MN/C

Electric field by a uniform disk:

The electric field due to a uniformly charged disk of radius R and uniform charged density σ, at a distance z above the center of the disk is given by:

$E = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})$

given that :

σ = 7.2×10⁻³C/m²

and R = 40cm

(i) The electric field at z = 5cm

$E = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})\\\\E = \frac{7.2\times10^{-3}}{2\8.85\times10^{-12}}(1-\frac{5}{\sqrt{5^2+40^2}})$

E = 356 MN/C

(ii) The electric field at z = 5cm

$E = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})\\\\E = \frac{7.2\times10^{-3}}{2\8.85\times10^{-12}}(1-\frac{10}{\sqrt{10^2+40^2}})$

E = 308 MN/C

(iii) The electric field at z = 5cm

$E = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})\\\\E = \frac{7.2\times10^{-3}}{2\8.85\times10^{-12}}(1-\frac{50}{\sqrt{50^2+40^2}})$

E = 89 MN/C

(iv) The electric field at z = 5cm

$E = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+R^2}})\\\\E = \frac{7.2\times10^{-3}}{2\8.85\times10^{-12}}(1-\frac{200}{\sqrt{200^2+40^2}})$

E = 7.7 MN/C

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