How Does Charge Density Affect Electric Field Inside a Cylindrical Insulator?

Then use the definition of work, W = qV, to find the required work to move the test charge from one point to the other. In summary, a long cylindrical insulator with a uniform charge density of 1.46uC/m3 and radius of 6cm has an electric field of 0.404N/C at a distance of 2cm and 0.0673N/C at 12cm. To move a test charge of q=0.086uC from 12cm to 2cm, the required work is 3.48 x 10^-5J. This can be found by integrating the electric field to find the potential and then using the definition of work.
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Homework Statement



A long cylindrical insulator has a uniformcharge density of 1.46uC/m3 and a radius of 6cm.a-What is the electric field inside the insulator at a distance of 2cm and 12cm? Answer should be in N/C.

b-How much work must you do to bring a q=0.086uC test charge from 12cm to 2cm? Answer in J.

Homework Equations


I have started with p=Q/V. With V= [tex]\pi[/tex]r2h

The Attempt at a Solution

...yet there is no h??
I know for r>R and r<R there are differences. I am lost. Please someone point me in the correct direction.
 
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  • #2
yet there is no h??

Well, it's there in the volume charge density that you've written, right? You have to construct an appropriate gaussian surface and apply Gauss's law to find the field. Can you proceed from here?
 
  • #3
Step 1. Write down Gauss's law: [tex]\oint \mathbf{E}\cdot d\mathbf{a} = \frac{Q}{\varepsilon_0}[/tex]
Step 2. Determine the direction of electric field (radially outwards) and the Gaussian surface you're going to use (a cylindrical shell).
Step 3. Find the total charge enclosed within your Gaussian surface. Hint: [tex]Q = \rho V[/tex].
Step 4. Do the math to find the field. The height [tex]h[/tex] will cancel on both sides of Gauss's law equation.

For part b, integrate the electric field to get the potential.
 

FAQ: How Does Charge Density Affect Electric Field Inside a Cylindrical Insulator?

1. What is a cylindrical Gaussian surface?

A cylindrical Gaussian surface is a hypothetical surface used in the mathematical analysis of electric fields. It is a cylindrical surface that encloses a certain area around a charge or a group of charges.

2. How is a cylindrical Gaussian surface used in Gauss's Law?

In Gauss's Law, a cylindrical Gaussian surface is used to calculate the electric flux through the surface. This can then be used to determine the electric field at a specific point outside the surface.

3. What are the properties of a cylindrical Gaussian surface?

A cylindrical Gaussian surface is an imaginary surface with the following properties: it is closed, symmetrical, and the electric field is perpendicular to the surface at every point. Additionally, the surface must enclose a certain area and be infinitely thin.

4. Can a cylindrical Gaussian surface be used for any charge distribution?

Yes, a cylindrical Gaussian surface can be used for any charge distribution as long as the surface encloses the entire charge or group of charges being analyzed. However, it is often most useful for cylindrical or spherical charge distributions.

5. How does the size of the cylindrical Gaussian surface affect the accuracy of the calculations?

The size of the cylindrical Gaussian surface can affect the accuracy of the calculations. A larger surface will give a more accurate result, but it may also be more difficult to calculate. A smaller surface may be easier to calculate, but it may also lead to a less accurate result.

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