Electric Field inside a cylinder

In summary: So when you calculate the area (which is the product of the circumference and the length), you only want to use the curved part. In summary, the electric field inside a long cylindrical insulator with uniform charge density can be found using the equation E = p*r/(2*ε0), where p is the charge density, r is the distance from the center, and ε0 is the permittivity of free space. This equation is derived by considering the flux through a Gaussian surface and only taking into account the curved part of the cylinder.
  • #1
orb123
3
0

Homework Statement



A long cylindrical insulator has a uniform charge density of 0.94 µC/m³, and a radius of 7cm (R). Find the electric field inside the insulator at a adistance of 4cm (r). Answer in units of N/C.


Homework Equations



Variables:

q= charge in the gaussian surface
R= Radius of cylinder
r= radius of gaussian surface of cylinder ; r<R
L= Lenght of gaussian surface of cylinder

Equations:

EA=q/ε0
V= (pi)R²L
V'= (pi)r²L
q=pV' volume of the gaussian surface of the cylinder
Q=pV; Q= charge, p=charge density
A=2(pi)RL = Area of cylinder (only the curved surface)

The Attempt at a Solution




Ok, I'm solving the gaussian surface of the cylinder. I know q<Q, and q=pV' = p{(pi)r²L}.
Then, I know p=Q/V by definition, so q=Q{(pi)r²L}/V.

Pluggin it into the first equation:

EA = Q{(pi)r²L}/{V(ε0)}
E{2(pi)RL}=Q{(pi)r²L}/{V(ε0)}

Solving:

E= Qr²/{2V(ε0)R} ; Then replacing V

E= Qr²/{2(pi)R²L(ε0)R} (??)

Why do I still have L there? I was not given a value for that, and it was supposed to be eliminated during solving the equations. And even if L wasn't there, the final solution would not be in units of N/C. What am I doing wrong?


Thanks!
 
Last edited:
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  • #2
Express the charge within your Gaussian surface as [itex]\rho V = \rho \pi r^2 L[/itex].
 
  • #3
I think I already did that. I replaced V' in q=pV' with (pi)r²L, but I think what I did after that was wrong, but I don't know why.

Thanks for your help!

orb123 said:
Ok, I'm solving the gaussian surface of the cylinder. I know q<Q, and q=pV' = p{(pi)r²L}.
Then, I know p=Q/V by definition, so q=Q{(pi)r²L}/V.
 
  • #4
orb123 said:
I think I already did that.
Then why do I see Qs everywhere? :wink:
 
  • #5
Yeah, thanks!

I also realized I was solving the Area wrong. Instead of A=2(pi)RL, should be A=2(pi)rL since I am analyzing the area of the gaussian surface of the cylinder, not the area of the entire cylinder.

Thanks again.
 
  • #6
Doc Al said:
Express the charge within your Gaussian surface as [itex]\rho V = \rho \pi r^2 L[/itex].

Hey Doc Al, this is Aleksey, I'm doing same problem with different values for variables

so: p = Q/V
where r< R

and I get the final answer as exactly pr/2ε0, but when I plug in my answer into the answer post in one of the programs we use in school for HW - it's wrong. What am I doing wrong?
 
  • #7
tua82912 said:
Hey Doc Al, this is Aleksey, I'm doing same problem with different values for variables

so: p = Q/V
where r< R

and I get the final answer as exactly pr/2ε0, but when I plug in my answer into the answer post in one of the programs we use in school for HW - it's wrong. What am I doing wrong?
Beats me. That answer looks correct.
 
  • #8
thanks, at least know on the right track from a Doc. Do you have a masters in Physics or you just like physics in general as an art and mentorship?
 
  • #9
I got it, my answer was right! Thanks so much for the opinion, i kept writing the formula correctly pr/2ε0 but was plugging into my calc r^2 all the time instead of r.
 
  • #10
How was the equation Pr/2ε0 derived?

I have a similar problem and the formula appears to be correct, but I'd like to know how you get to it

I know E = Q/ε0 = (PV)/(Aε0) = (P*pi*r^2)/(ε0(2*pi*r^2+2*pi*r*L))

I separated the equation into two and got P*L/(2*ε0)+P*r/(2*ε0). Why is it that only the latter part is the correct equation to use?

By the way, my problem involves a straight and long cylindrical charge cloud. We're given the radius and charge density, but no length.
 
  • #11
jhexp said:
I separated the equation into two and got P*L/(2*ε0)+P*r/(2*ε0). Why is it that only the latter part is the correct equation to use?
When calculating the flux through your Gaussian surface, only the curved side of the cylinder counts since the field is radial. The flux through the end pieces is zero since the field is perpendicular to those surfaces, so those areas don't count.
 

What is an electric field inside a cylinder?

An electric field inside a cylinder refers to the region surrounding the cylinder where electric charges experience a force due to the presence of an electric charge within the cylinder.

How is the electric field inside a cylinder calculated?

The electric field inside a cylinder can be calculated using the equation E = λ/2πε₀r, where E is the electric field, λ is the linear charge density, ε₀ is the permittivity of free space, and r is the distance from the center of the cylinder.

Does the electric field inside a cylinder depend on the radius of the cylinder?

Yes, the electric field inside a cylinder is inversely proportional to the radius of the cylinder. This means that as the radius increases, the electric field decreases.

Is the electric field inside a cylinder uniform?

No, the electric field inside a cylinder is not uniform. It is strongest at the edges of the cylinder and decreases as you move towards the center.

What factors can affect the electric field inside a cylinder?

The electric field inside a cylinder can be affected by the charge distribution within the cylinder, the radius of the cylinder, and the permittivity of the material surrounding the cylinder.

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