# Electric Potential Homework: Gauss's Law, Integrals, Bounds

• Aero6
In summary, the conversation discusses a problem involving a long coaxial cable with a volume charge density on the inner cylinder and a negative uniform surface charge density on the outer cylindrical shell. The goal is to find the electric field and potential difference in three different regions. The conversation also mentions using Gauss's law and the formula for surface area of a cylinder to solve the problem. Additionally, there is a discussion about the difference between using a strict "less than" sign and a "less than or equal to" sign in the problem.
Aero6

## Homework Statement

A long coaxial cable carries a volume charge density rho=alpha*s on the inner cylinder (radius a) and a uniform surface charge density on the outer cylindrical shell (radius b). This surface charge is negative and of just the right magnitude so that the cable as a whole is electrically neutral.
a)Find the electric field in each of the three regions:
s<a, a<s<b, s>b

b)find the potential difference in each of these regions with a refernce point at infinity

## Homework Equations

Gauss's law integral of E*da = Qencl/epsilon
V=-integral E*dl

## The Attempt at a Solution

b) a<s<b
I'm confused about integrating to find Qencl. Qencl=integral rho*dtao where dtao=s*ds*dtheta*dz, but when I set up the bounds on the integral for s, I don't understand which bounds I am supposed to include. Since there is a less than sign and not a less than or equal to sign when the problem says that s is greater than a and less than b, how is it okay to integrate so from a to an arbitrary distance that is less than b? Isn't this still including the distance a, which we shouldn't do because of the strict greater than sign? Also, how would this problem change if I was asked to find the electric field in the region: s is greater than OR equal to a and less than or equal to b?

Thank you

What shape is your Gaussian enclosure? I'm thinking of an infinitely long cylinder with the cable in the center. For the s > b, the total charge inside this is zero so an easy answer.
For a < s < b, there is charge and there may be some difficulty with charge and area being infinite but if you think "very long" instead of infinite, all that should cancel out. I don't even see the need for an integral - just the formula for the surface area of a cylinder.

The shape I'm using is a gaussian cylinder. Right, so the electric field for s> b is 0 because the coaxial cable is neutral, so we do not see any charge outside. The electric field for s<a can be solved using: E-field=Qencl/2*pi*r*L and Qencl can be found by using: Qencl=integral of rho*dtao, where dtao is rdr*dtheta*dz and plugging in Qencl into Gauss's law. I don't think I phrased my question correctly. in some coaxial problems we are told to find the electric field in the region between the cables where a<s<b where s is the radius of our Gaussian surface. In another problem (also dealing with a coaxial cable), the question has asked to find the electric field in the region where s is greater than OR EQUAL TO a and s is LESS THAN OR EQUAL TO b. Does the less than or equal to make a difference from a problem that does not ask less than or equal to? I know Gauss's law says that no points outside of our gaussian surface will act on the electric field inside our Gaussian surface, but for questions that ask less than or equal to, is it necessary to use 2 gaussian surfaces? one that integrates from a to s and one that integrates from s to b?

## 1. What is electric potential?

Electric potential is a measure of the potential energy that a charged object possesses at a certain point in space. It is measured in volts (V) and is a scalar quantity.

## 2. How is Gauss's Law used in calculating electric potential?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed charge. It can be used to calculate the electric potential at a point due to a distribution of charges.

## 3. What are integrals used for in electric potential calculations?

Integrals are used to calculate the electric potential due to a continuous charge distribution. By using the integral form of Gauss's Law, we can integrate over the charge density to find the electric potential at a specific point.

## 4. What do the bounds in electric potential integrals represent?

The bounds in electric potential integrals represent the limits of integration, or the boundaries of the region where the charge distribution is present. These bounds are important in determining the total electric potential at a specific point.

## 5. Can electric potential be negative?

Yes, electric potential can be negative. This indicates that the electric potential energy of a charged object at a certain point is decreasing as the object moves towards that point. It can also indicate a difference in electric potential between two points, where the point with the lower potential is considered to have negative potential.

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