(a) What is the magnitude E of theelectric field at a radial distance of r = 2.0

Question

(a) What is the magnitude E of theelectric field at a radial distance of r = 2.00R2?
1 N/C

(b) What is the direction of the electric field at that radialdistance?
2---Select---inwardoutward

(c) What is the magnitude E of the electric field at aradial distance of r = 7.50R1?
3 N/C

(d) What is the direction of the electric field at that radialdistance?
4---Select---inwardoutward

(e) What is the charge on the interior surface of the shell?
5 C

(f) What is the charge on the exterior surface of the shell?
6 C

Explanation / Answer

   for simplicity let us assume the charge densityof both the conducting cylinder and the shell are uniform and weneglect

    the firing effect, symmetry can be used toshow that the electric field is radial both between thecylinder and the shell

    and outside the shell, it is zero ofcourse inside the cylinder and inside the shell
(a)
   we take gaussian surface to be a cylinder oflength L coaxial with the given given cylinders and of largerradius r than
   either of them
   the flux through this surface is
    = 2 r L E
   where E is the magnitude of the field at thegaussian surface
   we can ignore any flux through the ends
   the charge enclosed by the gaussian surfaceis
   qenc = Q1 +Q2
           = -Q1
           =- 1.90 x 10-12 C
   so the gauss law yields to
   2 r o L E =qenc
   E = qenc / 2 oL R
      = ........ N / C
   |E| = ........... N / C
(c)  
   similar menner of part (a)
(e)
   we consider a cylindrical gaussian surface whoseradius places it with the shell itself the electric field is zeroat all points
    on the surface since any field within aconducting material would lead to current flow so the totalelectric flux through
    the gaussian surface is zero and the net chargewith in it is zero since the central rod has a charge Q1the inner surface
    of the shell nmust have chharge
   Qin = - Q1
         = - .........C
(f)
   since the shell is known to have total charge     
   Q2 = - 2.00 Q1
   it must have charge
   Qout = Q2 -Qin
           = - Q1
           = ........ C on its outer surface  

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