What is the maximum electric field magnitude between the cylinders?

[hidemi]

Homework Statement

Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conductor, with respect to the outer conductor, is +700V. What is the maximum electric field magnitude between the cylinders?

The given answer is 25,000 V/m.

Relevant Equations

E= (-kQ/r)⋅ dr

I used a couple ways to do this question, but I got neither correct. Can someone help, please? Thank you.

1. E= V/r = 700 / (60*10^-3) = 11667 (very far from the given answer)

2. E = (-kQ/r)⋅ dr
= kQ/r^2
= kQ/ [( 1/20/ 10^-3)^2 - (1/80/10^-3)^2]
(For this method, I stuck because I don't know kQ)

 

[Delta2]

Neither of the formula you use is correct for the electric field of such a system configuration. The correct formula is $$E=\frac{\lambda}{2\pi\epsilon_0}\frac{1}{r}$$ and $$V=\int Edr=\frac{\lambda}{2\pi\epsilon_0}\ln r$$ where $\lambda$ is the linear charge density (Coulomb/meter) of the inner cylindrical shell.
Now, first show me that you able to prove the above formulas (you have to use the integral form of Gauss's law to prove the first, the second is just the integral of the first) and then we can continue.

 

[hidemi]

Neither of the formula you use is correct for the electric field of such a system configuration. The correct formula is $$E=\frac{\lambda}{2\pi\epsilon_0}\frac{1}{r}$$ and $$V=\int Edr=\frac{\lambda}{2\pi\epsilon_0}\ln r$$ where $\lambda$ is the linear charge density (Coulomb/meter) of the inner cylindrical shell.
Now, first show me that you able to prove the above formulas (you have to use the integral form of Gauss's law to prove the first, the second is just the integral of the first) and then we can continue.

700 = λ/(2πε) *1.4
E = λ/(2πεr) = 700/(1.4*r) = 500/(2*10^-3)
I use 20*10^-3 meters to plug in 'r' and get the correct answer, but I"m not sure why.

 

[Delta2]

View attachment 282504
700 = λ/(2πε) *1.4
E = λ/(2πεr) = 700/(1.4*r) = 500/(2*10^-3)
I use 20*10^-3 meters to plug in 'r' and get the correct answer, but I"m not sure why.

The problem statement asks for the maximum electric field. For which value of r you think is the maximum as long as the electric field is given by the formula at post #2? But still you haven't proved the formula for the electric field.

 

[hidemi]

The problem statement asks for the maximum electric field. For which value of r you think is the maximum as long as the electric field is given by the formula at post #2? But still you haven't proved the formula for the electric field.

I think I has proved the formula for the electric field as my attachment in post #3(as below):

V=∫ (λ/2πε0r)⋅dr
= (λ/2πε0r)∫ (1/r)⋅dr
= (λ/2πε0r) ln r

Could you give me some more hints if it is not you want me to do which I attached in post #3 (as above)?

 

[Delta2]

That's the proof of the formula for the potential. The formula for the electric field $E=\frac{\lambda}{2\pi\epsilon_0}\frac{1}{r}$ you haven't prove it yet. To prove it you need to use Gauss's law and take as surface a cylinder with radius r in between the radius of 20mm and 80mm.

 

[hidemi]

That's the proof of the formula for the potential. The formula for the electric field $E=\frac{\lambda}{2\pi\epsilon_0}\frac{1}{r}$ you haven't prove it yet. To prove it you need to use Gauss's law and take as surface a cylinder with radius r in between the radius of 20mm and 80mm.

 

[Delta2]

Your first line is correct, but then i can't understand what you are doing in 2nd and 3rd lines. In 2nd line you have what it seems to me a line integral which simply is incompatible with the surface integral at line 1.
How can you proceed from line 1 to calculate that integral. Remember that our surface is a cylinder with radius r 20mm<r<80mm. Answer this question: How is the electric field magnitude and direction over this surface?

 

[hidemi]

Your first line is correct, but then i can't understand what you are doing in 2nd and 3rd lines. In 2nd line you have what it seems to me a line integral which simply is incompatible with the surface integral at line 1.
How can you proceed from line 1 to calculate that integral. Remember that our surface is a cylinder with radius r 20mm<r<80mm. Answer this question: How is the electric field magnitude and direction over this surface?

Is it correct?

 

[Delta2]

Yes that is correct well done!

 

[hidemi]

Yes that is correct well done

Thanks for confirming.
However, I am still unsure how to answer "What is the maximum electric field magnitude between the cylinders?"

As I calculated previously,
700 = λ/(2πε) *1.4
E = λ/(2πεr) = 700/(1.4*r) = 500/(2*10^-3)
I use 20*10^-3 meters to plug in 'r' and get the correct answer, but I"m not sure why.

 

[Delta2]

Well , r varies between 20 and 80mm. What can you say about the electric field as a function of r? when we increase r what happens to electric field? it increases or decreases? So the maximum value of E field is for low or high value of r?

 

[hidemi]

Well , r varies between 20 and 80mm. What can you say about the electric field as a function of r? when we increase r what happens to electric field? it increases or decreases? So the maximum value of E field is for low or high value of r?

Thanks for the hint. I got it!