# Use Gauss' Law to find E-field, tricky problem

cookiemnstr510510

## Homework Statement

A) use gauss's Law to determine the electric field at all values of radial distance (0<r<infinity) from the center of a non-uniformly charged cylinder that is very very long and lies along the x-axis. The cylinder carries excess charge per volume ρ=[a]r^2 (the [] are supposed to be absolute value) and has total radius,R.
B) a nonconductive sheet of paper lies in the xz-plane such that the bottom edge (width) of the paper infinitesimally close to the edge of the cylinder from part (A) and has its width parallel to the axis of the cylinder. The paper has width,W, along the x-axis, and length,L, along the z-axis, and carries a non-uniform linear charge density ([z]-R) (again the [] represent abs value). What is the total force acting on the sheet of paper in terms of the given constants (a,b,R,w,ε0 and L)
The problem is attached(IMG_4007. JPG) below.

gauss's Law

## The Attempt at a Solution

I have uploaded some of my interpretation of the problem (My interpretation.jpg)
My first thought is to look at a small segment of the cylinder with radius, r, thickness, dr, and charge, dQ.
we will find dQ by:
dQ=ρdV
dV=2πrLdr
dQ=ρ2πrLdr, however it is not uniform charge density. the problem states the cylinder carries excess charge per volume of ρ=ar2 (the "a" is in abs value)
so
dQ=ar22πrLdr
when we integrate dQ we will get Qenclosed within our gaussian surface.
Limits of integration for below integral are 0→R1 (R1 is the radius of my gaussian cylinder)
Q=∫dQ=∫ar22πrLdr=a2πL∫r3dr→→(aπLR14)/2
we now have our Q enclosed for gauss's Law
skipping a few steps of writing out gauss's law
E2πR1L=(aπLR14)/(2ε0)→ E=(aR13)/(4ε0)
For outside the cylinder we basically change limits when integrating dQ, Outer Gaussian cylinder radius R2
so our new limits of integration are 0→R
Qenclosed=a2πL∫r3d→→(aπLR4)/2
plugging into gauss's Law
E2πR2L=(aπLR4)/(2ε0)→E=(aR4)/(4R2ε0)

Now for part B.
I was talking with @TSny about a similar problem the other day, except this has a non-uniform charge density...I wasnt 100% good with the one with a uniform charge density. I need things broken down very basically. Not asking to give me the answer (as that is against the rules!!!), but definitely need some help.
Thanks in advanced, you guys/girls have been awesome so far!!!!!!!!!!!!!! :)

#### Attachments

• IMG_4007.jpg
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• My interpretation.jpg
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## Answers and Replies

Homework Helper
Gold Member
Don't use R as a variable. R is defined in the probem as the radius of the cylinder. Typically r is used as a variable. Use r in the integration to find Q, then you're done with integrations.

Once you have found Q you don't integrate again the way you did. Q is now a constant. Define a gaussian surface of radius r with Q as the free charge inside the cylinder. It looks like you computed Q correctly but I can't check your math.

Science Advisor
As post 2 says, you need to clearly distinguish r, R1 and R2 . Ultimately, what you need is the electric field as a function of the variable distamce r. Once you have got your symbols straightened out, you should be able to write the expressions for the field in terms of r, one expression for the field inside and one for outside the cylinder.

Homework Helper
Gold Member
I deleted my last post since it was misleading.
There is only one integration, to find the field from r=0 to r=R, which finds Q(r). At r=R you have Q(R). Beyond Q(R) there are no more integrations, just applying Gauss's law again.