Electric field intensity around a wire

[michael pranit]

Homework Statement

I am trying to calculate the surface charge density and electric field intensity around a copper wire. This is not a part of my assignment but out of curiosity. First, A current carrying wire of radius 2m, radius 0.5cm is at a height of 5cm above a ground plane, the ground plane is at a potential of 24V, the wire is at a potential of say 12V. How do I calculate the e.field intensity and surface charge density around the wire?

if a large current (like lightning) of 10KA having charge of 26C, and frequency 100Hz flows through the wire. what would be the charge on the wire and How would it effect the potential since (V=I*R)?

Homework Equations

The charge density ρ=(q/A)
Efield intensity E=λ/(2∏ε0*R)
where R is the radius of the gaussian surface, λ is the charge per unit length

The Attempt at a Solution

Attributing the wire to be a cylinder; the surface area would be A=2∏rh+2∏(r^2) = 0.0629 m2.
here h = length of the wire.
To get the value of q for surface charge density.(I know I'm wrong)
q = I*t
t= 1/f = 1/100 = 10ms; I = 10KA
q = 100C
charge density ρ= 1.5K C/m2 ??

to solve E.field intensity, I considered a gaussian surface with the distance between the wire and ground plane as the radius R and the length of the gaussian surface equal to the length of the cylinder
if the lightning carriers charges of 26C then λ= 26/2 = 13 (assuming uniform distribution of charge on surface)
E = 13/(2∏*ε0*2)
E = 0.1*10^-12 V/m??

I'm Extremely confused. Any help will be really appreciated.

 

[KataKoniK]

Hello there,

Calculating the surface charge density and electric field intensity around a copper wire is a very interesting problem. Let's break it down step by step.

First, to calculate the surface charge density, we use the equation ρ = q/A, where q is the total charge and A is the surface area. In this case, the wire is at a potential of 12V and the ground plane is at a potential of 24V. The potential difference between them is 12V. We can use the equation V = IR to find the current I, which is equal to 12V/2Π = 6V/Π. Now, we know the current and the time (t = 10ms) so we can find the total charge by multiplying them: q = It = 6V/Π * 10ms = 60μC.

Next, we need to find the surface area of the wire. You have correctly calculated the surface area as A = 2Πrh + 2Πr^2. Plugging in the values, we get A = 2Π*2m*5cm + 2Π*(0.5cm)^2 = 0.0629m^2.

Now, we can calculate the surface charge density as ρ = q/A = 60μC/0.0629m^2 = 952.5μC/m^2. This is the charge per unit area on the surface of the wire.

Moving on to the electric field intensity, we can use the equation E = λ/(2Πε0r), where λ is the charge per unit length and r is the distance from the wire. In this case, we can take the distance from the wire to the ground plane as the radius r. We already know the charge per unit length (λ = q/h = 60μC/2m = 30μC/m) and the value of ε0 (8.85*10^-12 F/m). Plugging in the values, we get E = 30μC/m/(2Π*8.85*10^-12 F/m) = 0.00107 V/m. This is the electric field intensity at a distance of 5cm from the wire.

Now, let's consider the scenario where a large current of 10KA flows through the wire with a frequency of