Cylindrical Gaussian surface

by PhysicsRetard
Tags: cylindrical, gaussian, surface
PhysicsRetard is offline
Jun30-08, 10:56 PM
P: 1
1. The problem statement, all variables and given/known data

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.

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

3. 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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siddharth is offline
Jul1-08, 04:18 AM
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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?
Irid is offline
Jul2-08, 10:54 AM
P: 208
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.

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