# Voltage of an infinite cylinder with nontrivial reference point

• programmer1813
In summary, the question asks to find the electric potential due to an infinitely long cylinder with uniform charge density ρ and radius R, using V(r = 2R) = 0 as the reference point. The electric fields for r < R and r > R are given by ##\frac{(ρr)}{2ε_{0}}## and ##\frac{(ρR^{2})}{(2ε_{0}r)}## respectively. To find the potential, one must integrate the electric fields. For r < R, the limits of integration are from r to R and from R to 2R. For r > R, the limit is from r to 2R. The resulting equations for V(r <
programmer1813

## Homework Statement

Find V(r), the electric potential due to an infinitely long cylinder with uniform charge density ρ and radius R.
Use V(r = 2R) = 0 as your reference point.

## Homework Equations

E at r < R = ##\frac{(ρr)}{2ε_{0}}##
E at r > R = ##\frac{(ρR^{2})}{(2ε_{0}r)}##

## The Attempt at a Solution

This is basically simple. Just integrate the electric fields to get voltage. The only problem is figuring out the limits of integration. I BELIEVE that if you have a nontrivial reference point outside of a cylinder, you will need to integrate from r to 2R for both inside and outside. So I did this:

$$V(r < R) = \int_r^R \frac{\rho r}{2ε_{0}}dr + \int_R^{2R} \frac{\rho r}{2\epsilon_{0}}dr = \frac{\rho}{4\epsilon_{0}}(4R^{2} - r^{2})$$

$$V(r > R) = \int_r^{2R} \frac{\rho R^{2}}{2\epsilon_{0}r}dr = \frac{\rho R^{2}}{2\epsilon_{0}}ln(\frac{2R}{r})$$

Does this look right?

Assuming your field calculations are correct, these look right.

I'm not following the logic behind your equations.
For r > R, the field is ##E = \frac{\rho r}{2\pi\epsilon_0 R^2}##, yes?
So for r > R, isn't the potential ##\int_{s=r}^{2R}\frac{\rho s}{2\pi\epsilon_0 R^2}.ds##?

## 1. What is the voltage of an infinite cylinder with a nontrivial reference point?

The voltage of an infinite cylinder with a nontrivial reference point depends on the distance from the reference point to the surface of the cylinder. It can be calculated using the formula V = kλ/2πε, where k is a constant, λ is the charge density of the cylinder, and ε is the permittivity of the surrounding medium.

## 2. How does the voltage change as the distance from the reference point to the surface of the cylinder increases?

As the distance increases, the voltage decreases exponentially. This is because the electric field created by the cylinder weakens with distance, resulting in a lower voltage at farther distances from the reference point.

## 3. What is a nontrivial reference point?

A nontrivial reference point is a point located outside of the infinite cylinder and not on its axis. It is used as a reference point for calculating the voltage of the cylinder.

## 4. Can the voltage of an infinite cylinder with a nontrivial reference point be negative?

Yes, the voltage can be negative if the direction of the electric field is opposite to the direction of the reference point. This can occur if the charge density of the cylinder is negative and the reference point is located closer to the cylinder than the surface.

## 5. What are the applications of understanding the voltage of an infinite cylinder with a nontrivial reference point?

Understanding the voltage of an infinite cylinder with a nontrivial reference point is important in various fields such as electromagnetism and engineering. It can help in the design and optimization of electrical devices, as well as in understanding the behavior of electric fields in different situations.

Replies
11
Views
420
Replies
6
Views
569
Replies
4
Views
950
Replies
11
Views
2K
Replies
3
Views
2K
Replies
6
Views
2K
Replies
13
Views
1K
Replies
36
Views
891
Replies
12
Views
2K
Replies
11
Views
3K