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Electric Field Question 
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#1
Sep604, 02:20 PM

P: 33

Consider a long uniformly charged, cylindrical insulator of radius R with charge density 1.1 microcoulombs/m^3. (The volume of a cylinder with radius r and length l is V = pi*r^2*l)
What is the magnitude of the electric field inside the insulator at a distance 2.7 cm from the axis (2.7 cm < R)? Answer in units of N/C.  The axis they are referring to in the problem runs through the cylinder from top to bottom... i dont really know where to go with this problem. any pointers/tips/starting points would be great. 


#2
Sep604, 03:06 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,533

The electrical field from a single point charge q is the vector qr/r^{3} where r is the vector from the charge to the point and r is the length of that vector. (The r in the numerator just gives the direction of the vector. The cube, rather than a square, in the denominator is to cancel the length of that vector.)
Set up a cylindrical coordinate system with origin at the center of one base and positive zaxis along the axis of the cylinder. The "differential of volume" in cylindrical coordinates is r dr dθdz and the "differential of charge" is ρr drdθdz where ρ is the charge density. Integrate (ρr/rho^{3}) rdrdθdz (r is the vector from the given point (x,y,z) to the point in the cylinder and r is the distance from the origin to to the point in the cylinder) over the cylinder. 


#3
Sep704, 04:46 PM

Sci Advisor
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P: 2,895

The simplest way to answer this is to use Gauss' law. Have you learned this? Set up a cylindrical gaussian surface or radius r< R and length L (with its axis coincident with the axis of the real cylinder). The net flux through your gaussian surface will be the magnitude of the E field at a distance r times the surface area of the curved side of the gaussian surface, namely [itex] \Phi = E 2 \pi r L [/itex]. On the other hand, the net flux is also the total charged contained inside your gaussian surface divided by [itex] \epsilon_0 [/itex], according to Gauss' law, i.e. [itex] \Phi = q_{in}/ \epsilon_0[/itex]. The charge contained inside your gaussian surface is [itex] q_{in} = \rho \times \pi r^2 L [/itex]. Now set the two expressions for the flux equal to one another (the length L of your gaussian surface will cancel out) and solve for E. Sub in the values for r (the 2.7 cm), [itex] \rho[/itex] and you're done. Pat 


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