Question

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a distance a = 0.75 m from the center of the disk and perpendicular to the face of the disk.
(1) Calculate the electric field at point P in units of kilonewtons per coulomb.
(2) Write an equation for the electric field at point P when a< (3) Write an equation for the electric field at point P when a>>R. When a>>R the disk appears as a single point.

Answer 1

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

• The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  $E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

Answer 2

1)  The electric field at point P is 1408.685 KN/C

2)  An equation for the electric field at point P when a<  is $E = \sigma / 2*e_0$

3) An equation for the electric field at point P when a>>R is  $E=(k*\delta*\sigma/a^2)$

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = $8.99*10^9$

What is Electric Field Strength?

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E= 2*\pi *k*\sigma*(1-\frac{a}{\sqrt{a^2+R^2} } )$                    

Part (a)

Electric Field strength at point P: a = 0.75 m

$E= 2*\pi *k*\sigma*(1-\frac{a}{\sqrt{a^2+R^2} } )\\\\E= 2*\pi *8.99*10^9*175*10^{-6}*(1-\frac{0.75}{\sqrt{0.75^2+0.45^2} } )\\\\E=1408.685KN/C$

Part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E= 2*\pi *k*\sigma*(1-\frac{a/R}{\sqrt{a^2/R^2+1} } )\\\\E= 2*\pi *k*\sigma*(1-\frac{0}{\sqrt{0+1} } )\\\\E= 2*\pi *k*\sigma\\\\E=\sigma/2*e_0$

Part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

$E = k*\delta*\pi*R^2 / a^2$

Since, R << a, Surface area = δ*pi

Hence,

$E=(k*\delta*\sigma/a^2)$

Find more information about Electric field strength here:

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