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

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SUMMARY

The discussion focuses on calculating the electric field inside a long cylindrical insulator with a uniform charge density of 1.46 µC/m³ and a radius of 6 cm. The electric field at distances of 2 cm and 12 cm from the center is determined using Gauss's law, specifically the equation \(\oint \mathbf{E}\cdot d\mathbf{a} = \frac{Q}{\varepsilon_0}\). Additionally, the work required to move a test charge of 0.086 µC from 12 cm to 2 cm is calculated by integrating the electric field to find the potential difference.

PREREQUISITES
  • Understanding of Gauss's law in electrostatics
  • Familiarity with cylindrical coordinates and charge density concepts
  • Knowledge of electric field calculations
  • Basic integration techniques for calculating work done
NEXT STEPS
  • Study the application of Gauss's law in different geometries
  • Learn about electric field calculations for non-uniform charge distributions
  • Explore the concept of electric potential and its relationship to electric fields
  • Practice problems involving work done on charges in electric fields
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields and potential in cylindrical geometries.

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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= \pir2h

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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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?
 
Step 1. Write down Gauss's law: \oint \mathbf{E}\cdot d\mathbf{a} = \frac{Q}{\varepsilon_0}
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: Q = \rho V.
Step 4. Do the math to find the field. The height h 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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