# Homework Help: Voltage of cylinder of radius R

1. Mar 5, 2014

### TwinCamGTS

ε1. The problem statement, all variables and given/known data
A very long solid cylinder of radius R = 6 cm has:
-uniform volume charge density ρ = +7 µC/m3.
-linear charge density λ = 7.9168E-8 C/m

Outside: What is the electric field at a radial distance of 7 cm from the axis of the cylinder?
Inside: What is the electric field at a radial distance of 5 cm from the axis of the cylinder?

Voltage: Setting V=0 at r=3R, find the voltage at the following locations:
V (r = 0)
V (r = R/2)
V (r = R)
V (r = 2R)
V (r = 4R)

2. Relevant equations

using gauss law to find the electric field inside and outside the cylinder

inside the cylinder E = ( ρ*r )/(2*εo)

outside the cylinder E = ( λ )/(2*∏*r*εo)

Voltage is just integration of E, so V=-∫Edr

3. The attempt at a solution

I find the electric field from the axis 0m to 0.07m using E = ( λ )/(2*∏*r*εo), i got 20338.948 N/C

I find the electric field from the axis 0m to 0.05m using E = ( ρ*r )/(2*εo), i got 19773.9484 N/C

The problem i have is to find the voltage from the asked distance to distance r=3R
we set V at 3R = 0.

i do the integration on E = ( ρ*r )/(2*εo) from r=0 to r=R to find the Voltage inside the cylinder

then do integration on E = ( λ )/(2*∏*r*εo) from r=R to r=3R to find the voltage outside the cylinder.

when it ask find the voltage at r=0, it means that i have to find the ΔV, which is from
V(r=0) to v(r=3R), then i do separate integral on that two different E equation, then add them up. Since V at r=3R is 0, then i pretty much ignored it during the integration. but i got it wrong. how am i supposed to do this. i have try many method and i think this is the correctway to do it. can someone help me out or explaining it to me? Thank you for your time and reply

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2. Mar 5, 2014

### jackarms

For finding the potential at r = 0, you're right in having to add up the two integrals, one from r = 3R to r = R, then another from r = R to r = 0, but in your work there's only one. What happened to the other?

3. Mar 5, 2014

### TwinCamGTS

yeah, thats what i thought but the webassign said my answer is wrong, so i wonder what is the way to do it then?

so you say the method im doing it is correct isnt it?

Or is be because the equation that i use is wrong?
when r=R, do i use equation of r gauss < R cylinder or r gauss > R cylinder?

if that is the way to do it, then might be the way i setting up the equation for the integral is wrong.

what do you thnk?

4. Mar 5, 2014

### jackarms

Yes, that method should be correct. Think about it like this:
$$\Delta V = V(r) - V(3R)$$
So this means that
$$V(r) = V(3R) + \Delta V$$
You know that V(3R) = 0, and also that you start integrating from 3R. So say to find the potential at r = 0:
$$V(r = 0) = 0 - \int^{R}_{3R}E_{outside}dr - \int^{0}_{R}E_{inside}dr$$
I think that's what you were doing -- you may have been a little off in the bounds or what field to use perhaps. And when r = R, you don't really pick between the fields -- R is when the field changes, so it's more that you switch the fields between the two integrals.

5. Mar 5, 2014

### TwinCamGTS

i dont know what to do.
I just do the easiest thing first, which is find the voltage V at r=R
based on the integration, i have 2 answers.
either 711.864 (integrated from r=0 to r=R) or 4005.527 (integrated from r=3R to r=R)
i put down either answer, and its wrong....

i find out that its weird, based on the graph of the voltage over radius,
the voltage from r=0 to r=R should be greater than from r=R to r=3R.
but based on the integral value, its the opposite.

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6. Mar 5, 2014

### jackarms

Okay, so you evaluated both integrals, correct? But the answer isn't either of those -- look at the expression I posted. Those integrals are added together, right? Also, talking about the voltage at R is misleading -- what that means is the potential with respect to zero (V(3R)), so only the value of V(R) - V(3R) is meaningful. V(0) - V(R) means the potential difference between 0 and R, but says nothing about the potential at R.

7. Mar 5, 2014

### TwinCamGTS

I finally got it right. Apparently i was stupid to cross out the value during the integration just because they set the V=0 at r=3R. so i ended up get everything wrong.
thank you for your help :D
i appreaciate it