Gaussian Cylinder Electric Field and Net Charge Calculation

[Doc Z]

Homework Statement

A cylindrical shell of radius 7.00 cm and length 260 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.0 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Find values for the following.

(a) the net charge on the shell

(b)the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell

Homework Equations

E = Q/(4π r² εo)

The Attempt at a Solution

This problem is giving me such a headache. I thought you just have to rearrange the above equation to solve for Q as that is the total charge but I'm wrong. I don't understand what's going on conceptually with all these gaussian problems. I looked at online and book figures and it's just not clicking yet. PLEASE help!

 

[king vitamin]

the equation you're using is not Gauss' Law - it's the electric field due to a point charge or spherical distribution of charge.

Gauss' Law:

$$\int{E dA} = \frac{Q}{\epsilon o}$$

The integral might look complicated, but for basic problems you can rewrite it for the following symmetries:

Spherical: E * 4 π R²
Cylindrical: E * 2 π R L
Planar: E * A

 

[Doc Z]

Thank you so much!

I am so confused with all this electric stuff...it's like a foreign language to me. It's the first time I'm completely lost in my Physics class.

 

[Doc Z]

Ugh still a little confused.

So I use E * 4pie R^2 --> 36000 * 4pie (.07^2)

How do I find the net charge though?

 

[Doc Z]

So the equation I just used helps me find the net electric flux which equals the net charge/e0.

So to find the net electric charge, I should multiply epsilon0 by the net electric flux...but I get a wrong answer.

 

[king vitamin]

What expression did you get for the net electric flux through a cylinder?

 

[Doc Z]

That would be E*4pieR^2 right?

and my answer to that was 2216.708 using a radius of .07m. Then multiplying that number by epsilon0 should give me the total charge right?

 

[king vitamin]

No, that's the flux through a sphere. The area of the length side of a cylinder if 2*pi*r*l.

 

[Doc Z]

Ok, so q = (E*2*pi*r*l) * (epsilon0)

That's q= 36000*2*pi*.07*.260 * (8.88542 x 10^-12) = 3.66 x 10^-8.

It's still the wrong answer.

 

[king vitamin]

The electric field you're given is located at a distance of 20 cm from the radius of the cylinder, not 7 cm! Also, 260 cm = 2.6 m.

I think you're not picturing flux correctly - what you're doing is imagining a hypothetical cylinder wrapped around your charged cylinder at a certain distance (in this case 20 cm from the axis), and given a constant flux (E field times area of the length of the cylinder) Gauss' Law is telling you the amount of charge contained in your imaginary cylinder (which, by the problem statement, is the charge on the real cylinder).

 

[Doc Z]

Ah yes you are absolutely correct. Your description is perfect! So am I understanding that the net charge I am solving for in this problem is that of the imaginary cylinder which is the SAME as the real cylinder?

But...even with the mistakes I made, I'm still getting the wrong answer for some reason. I replaced .07 with .20m and I changed .26 to 2.6m.

q= 36000*2*pi*.20*2.60 * (8.88542 x 10^-12) = 1.05 * 10^-6.

edit: nevermind, I realized I had the wrong units. This is correct. I'm sorry I haven't had any sleep and I'm a little off lol

 

[Doc Z]

I don't understand why the answer to part b would be 0 though.

It's asking to find the electric field.

That means E= q/(epsilon0*2*pi*R*l)

The problem states that it is 4.0 cm from the axis which means that r=.04m.

Using the net charge found from part a (1.05 * 10^-6)/(epsilon0 * 2 * pi * .04m * 2.6m)...this doesn't give me 0 though.

 

[borgwal]

I don't understand why the answer to part b would be 0 though.

It's asking to find the electric field.

That means E= q/(epsilon0*2*pi*R*l)

The problem states that it is 4.0 cm from the axis which means that r=.04m.

Using the net charge found from part a (1.05 * 10^-6)/(epsilon0 * 2 * pi * .04m * 2.6m)...this doesn't give me 0 though.

The "q" in that formula is the total charge inside an imaginary cylinder of radius 4 cm. And that charge is...?

 

[Doc Z]

Apparently 0 lol. But I don't understand why...it's the same concept as part a isn't it? Why is the net charge 0 in this case?

 

[borgwal]

Apparently 0 lol. But I don't understand why...it's the same concept as part a isn't it? Why is the net charge 0 in this case?

How much charge is there inside the radius of 4cm? Where is the charge you calculated in part a located?

 

[Doc Z]

That charge is located 20cm away from the axis.

Is it because 4cm is within the 7cm of the real cylinder, thus making it 0 since the inside charge of a closed surface is always 0?

 

[borgwal]

That charge is located 20cm away from the axis.

Is it because 4cm is within the 7cm of the real cylinder, thus making it 0 since the inside charge of a closed surface is always 0?

The charge is located 7cm away from the axis, not 20cm. But in either case, yes, inside a radius of 4cm there is no charge!

The inside charge of a closed surface can be anything, by the way. But in THIS case, it's zero for a cylinder of radius of 4cm.

 

[Doc Z]

So anything up to 7cm would be 0, anything above would be the equation I used earlier?

 

[borgwal]

Exactly!

 

[Doc Z]

Thank you all for your help! Seriously, I cannot tell you how grateful I am. This forum is really appreciated, especially with members like you!

Thanks again.