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Electric field / net charge on a cylindrical shell

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Homework Statement



A cylindrical shell of radius 8.1 cm and length 247 cm has its charge density uniformly distributed on its surface. The electric field intensity at a point 20.3 cm radially outward from its axis (measured from the midpoint of the shell ) is 41800 N/C.
Given: ke =8.99×10^9 N·m^2 /C^2 . What is the net charge on the shell? Answer in units of C.

Homework Equations



Gauss's Law

The Attempt at a Solution



I've been using E=[tex]\lambda[/tex]/2[tex]\pi[/tex][tex]\epsilon[/tex]r
I used the radius from the midpoint (20.3 cm) in place of the r in the equation and used the given E to solve for [tex]\lambda[/tex]. Then I found the area of the cylindrical shell by doing 2.47*2[tex]\pi[/tex]*.081+2[tex]\pi[/tex]*.081 and multiplied that by the [tex]\lambda[/tex]. I ended up with an answer of 6.1288E-7 the last time I did it.

Is the r supposed to be distance from the surface, maybe? or am I using the completely wrong equation?

Thank you!
 
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Answers and Replies

  • #2
collinsmark
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Are you supposed to approximate the answer using Gauss' law, or are you supposed to find the exact answer through detailed integration? (in other words, which section of your textbook is this problem given? And does your course require lots of calculus?)

Either way, Go back to the original definition of Gauss' law, that gives you a relationship in terms of [tex] Q_{enc} / \epsilon _0 [/tex]. That way, you don't need to drag [tex] \lambda [/tex] into things (the question asks for the total charge, not the charge per unit length).

If the particular problem is really looking for an approximation, then that can be done fairly easily. You don't know what the electric field is at the end-caps of the Gaussian surface, but I'm guessing that you are supposed to assume that the electric field component, parallel to the end-caps, contribution is minimal. In other words, I'm guessing that you don't need to include the end caps of the Gaussian surface into the equation.

On the other hand, if you need the exact solution, you will need to integrate over the entire surface, without using the approximation that that the electric field is constant over most of the Gaussian cylinder (I'm guessing that you probably are allowed to make the approximation, but that's just my guess).

And no, r is the center of the cylindrical shell, which also the center of the Gaussian surface too.
 
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  • #3
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We're doing Gauss's law, so I'm assuming the approximation is okay.

I did EA=Q/[tex]\epsilon[/tex] and set the A to h*2[tex]\pi[/tex]r
(r being the r of the Gaussian surface, .203m)

It wasn't right...what am I doing wrong?
 
  • #4
ideasrule
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h is the length of the cylinder, right? If so, then your equation is right.
 
  • #5
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Ya, sorry.
Okay thanks! I'm not sure why my answer's coming out wrong...I guess I'll just show my teacher my work and see what she says haha
 

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