Cylindrical Shells and Gauss' Law

  • #1

Homework Statement



An infinitely long cylindrical shell of radius 6.0 cm carries a uniform surface charge density sigma = 12 nC/m^2. The electric field at r = 5.9 cm is approximately


a.0.81 kN/C

b.zero.

c.1.3 kN/C.

d.12 kN/C.

e.0.56 kN/C.



Homework Equations



See below.

The Attempt at a Solution



Will the correct choice be b. 0? I am assuming that this is a conducting cylindrical shell. When r < R, E will be 0.




Homework Statement



An infinitely long cylinder of radius 4.0 cm carries a uniform volume charge density rho = 200 nC/m^3. What is the electric field at r = 8.0 cm?


a.0.23 kN/C

b.0.11 kN/C

c.57 kN/C

d.0.44 kN/C

e.zero



Homework Equations



electric flux = Integral [E*dA] = Q_enclosed/epsilon_0

The Attempt at a Solution



I sloshed through this one because I mixed up the radii (r and R) at first. Did I eventually get the correct answer?

Q_enclosed = rho*2*pi*R^2*l

flux = E*2*pi*r*l

E*2*pi*r*l = (rho*2*pi*R^2*l)/(epsilon_0)

E = (rho*R^2)/(r*epsilon_0) = (200*10^-9 c/m^3)(0.04 m)^2/[0.08 m*8.85*10^-12] = 452 N/C??



Thanks.
 

Answers and Replies

  • #2
Doc Al
Mentor
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Homework Statement



An infinitely long cylindrical shell of radius 6.0 cm carries a uniform surface charge density sigma = 12 nC/m^2. The electric field at r = 5.9 cm is approximately


a.0.81 kN/C

b.zero.

c.1.3 kN/C.

d.12 kN/C.

e.0.56 kN/C.



Homework Equations



See below.

The Attempt at a Solution



Will the correct choice be b. 0? I am assuming that this is a conducting cylindrical shell. When r < R, E will be 0.
No reason to think the shell has any thickness or that it's a conductor. Nonetheless, since the charge is on the surface of the shell (and we presume no other charge exists) the field will be zero for r < R. (Since the charge enclosed within any cylindrical shell with such a radius would be zero.)


Homework Statement



An infinitely long cylinder of radius 4.0 cm carries a uniform volume charge density rho = 200 nC/m^3. What is the electric field at r = 8.0 cm?


a.0.23 kN/C

b.0.11 kN/C

c.57 kN/C

d.0.44 kN/C

e.zero



Homework Equations



electric flux = Integral [E*dA] = Q_enclosed/epsilon_0

The Attempt at a Solution



I sloshed through this one because I mixed up the radii (r and R) at first. Did I eventually get the correct answer?

Q_enclosed = rho*2*pi*R^2*l

flux = E*2*pi*r*l

E*2*pi*r*l = (rho*2*pi*R^2*l)/(epsilon_0)

E = (rho*R^2)/(r*epsilon_0) = (200*10^-9 c/m^3)(0.04 m)^2/[0.08 m*8.85*10^-12] = 452 N/C??
I didn't check your arithmetic, but your method is correct.
 

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