# Calculating electric field strength

## Homework Statement

A cylinder of radius R= cm 1.2 and length L= 51 cm has a charge Q=2.3 μC spread uniformly along its surface (and not on its flat ends).
a) Calculate the electric field strength a distance d=4 mm from the cylinder’s surface (not near either end)
b)Calculate the electric field strength a distance D=24 m from the rod

## Homework Equations

E= kQ/r^2

Surface area of cylinder without faces = 2πrL

Volume of cylinder = πr2L

pr/(2ε°) = E

## The Attempt at a Solution

My grade 12 teacher never taught us electricity so to lightly put it - I'm screwed (so please bare with me and anything I'm misunderstanding)

if my point charge is d=4 mm from the cylinder then from the CENTER it is actually R+d distance away. (similarly, R+D distance for part b).

Now, I know I don't need to do any integration as the charge is spread uniformly along the surface.

This is where I'm not sure what to do. I know that the surface area of the cylinder without its ends is 2πRL, which is the surface that the charge covers. Do I divide Q by the surface area? or do I find volume and do I create my "Gaussian surface" ie, a cylinder and use the radius R+d or R+D instead of R to find the volume? I don't know where to go from here, if someone could explain this, and some of the physics behind it, it might help clear this up for me for good.

Thank you!

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kuruman
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The general solution to this problem is quite complicated. This problem is actually asking you to approximate the solution. If you look at the distances where you need to find the field, you will see that in (a) you are very close to the surface. What does a cylinder look like at distances much smaller than its radius? Hint: What does the Earth look like at heights above its surface that are much smaller than the Earth's radius? In part (b) you are very far from the cylinder. What does the cylinder look like at distances much larger than its radius or length?

If you look at the distances where you need to find the field, you will see that in (a) you are very close to the surface. What does a cylinder look like at distances much smaller than its radius? Hint: What does the Earth look like at heights above its surface that are much smaller than the Earth's radius? In part (b) you are very far from the cylinder. What does the cylinder look like at distances much larger than its radius or length?[/QUOTE]

OHHHHH when you're really far away from the cylinder- the cylinder's dimensions no longer matter, it is like a point the further you get from it.

When you're really close to the cylinder, uh I'm gonna say it looks like the area of a rectangle but I'm not sure if that's right.

haruspex
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When you're really close to the cylinder, uh I'm gonna say it looks like the area of a rectangle but I'm not sure if that's right.
No, but the field does look like that of an infinite cylinder. What directions are the field lines in?

I get part b, for part a im not sure if I'm exactly correct but everything in my head is telling me to find the linear density (and integrate even though I don't need to do that now since I have an already integrated formula ready to use).

kuruman
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... my head is telling me to find the linear density ...
My head is telling me that when I am very close to a cylinder, the cylinder looks like a plane rather than a line. Which head do you think gives a closer approximation to the actual picture?

My head is telling me that when I am very close to a cylinder, the cylinder looks like a plane rather than a line. Which head do you think gives a closer approximation to the actual picture?
definitely yours but I'm not sure how to work with that.

kuruman
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What is the expression for the electric field due to a plane of uniform charge density σ?

haruspex
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What is the expression for the electric field due to a plane of uniform charge density σ?
Two problems with this approach.
The flux from an infinite plane escapes both sides, so you get a factor /2.
4mm is not that small compared to the 12mm radius of the cylinder.
Need to treat it as an infinite cylinder, not a plane.

What is the expression for the electric field due to a plane of uniform charge density σ?
we haven't learnt that yet :(

kuruman
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The flux from an infinite plane escapes both sides, so you get a factor /2.
I disagree. I think it would be more correct to use σ = Q/(2πRL) for the surface charge density of the "plane".
4mm is not that small compared to the 12mm radius of the cylinder.
Need to treat it as an infinite cylinder, not a plane.
I agree. For some reason I fixed in my head that a plane is a more appropriate approximation. The fractional difference between the two approximations is $\frac{E_{line}-E_{plane}}{E_{line}} =-\frac{d}{R}=- \frac{1}{3}.$

haruspex
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I disagree. I think it would be more correct to use σ = Q/(2πRL) for the surface charge density of the "plane".
For charge density σ, the field from an infinite plane is σ/(2ε0). The field just outside an infinite cylinder is σ/(ε0).

For a plane, half the flux lines emanate each side. Near a cylinder's surface they only emanate out from the cylinder.

kuruman
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For charge density σ, the field from an infinite plane is σ/(2ε0). The field just outside an infinite cylinder is σ/(ε0).
For a plane, half the flux lines emanate each side. Near a cylinder's surface they only emanate out from the cylinder.
All that is true. But what is σ in this case? Clearly, $\sigma = Q/(2\pi RL)$. In the "just outside" planar approximation, the electric field is $E=\sigma/ \epsilon_0 = Q/(2\pi \epsilon_0 RL).$ Compare with the "line of charge" expression which I will not write out because OP has not reached it yet.

haruspex
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In the "just outside" planar approximation, the electric field is E=σ/ϵ0
No, it is E=σ/2ϵ0. This is my point.

kuruman
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And here is my point. You have to agree that the surface charge density is $\sigma = Q/(2\pi RL)$. If I construct the standard Gaussian pillbox with one face just outside the cylinder and the other just inside, the total flux is $E dA$ through the outside and zero through the inside face. The enclosed charge is $\sigma dA = Q dA/(2\pi RL)$ so that $E=Q /(2\pi RL).$ If, instead, the pillbox extends through and out the opposite surface of the cylinder, the flux through it is doubled but so is the enclosed charge and the electric field is the same as before.

The expression E=σ/2ϵ0 is the field on either side of an infinite plane that is infinitely thin. The result follows from Gauss's Law. Now consider an infinite conducting plane of thickness d with surface charge density σ. The field on either side of the plane is σ/ε0. This result also follows from Gauss's Law. It is not in contradiction with the previous case because if you shrink the thickness of the conducting plane to zero, you get twice the charge density distributed over the plane so E = 2σ/(2ε0)=σ/ε0.

haruspex
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If I construct the standard Gaussian pillbox with one face just outside the cylinder and the other just inside, the total flux is EdA through the outside and zero through the inside face.
Sure, but that is not the same as treating it as merely a charged plane. Such would not be zero through the inside face.

My concern is that telling the OP to treat it as a charged plane will likely lead to the wrong formula being used.

kuruman
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OK, perhaps charged conducting plane would be more appropriate.

haruspex