Cylindrical Shells and Gauss' Law

Click For Summary
SUMMARY

The discussion centers on calculating the electric field generated by an infinitely long cylindrical shell and an infinitely long cylinder. For the cylindrical shell with a radius of 6.0 cm and a surface charge density of 12 nC/m², the electric field at a radius of 5.9 cm is definitively zero, as the charge is uniformly distributed on the surface. In contrast, for the cylinder with a radius of 4.0 cm and a volume charge density of 200 nC/m³, the electric field at a radius of 8.0 cm is calculated using the formula E = (rho*R²)/(r*epsilon_0), resulting in an electric field of approximately 452 N/C.

PREREQUISITES
  • Understanding of Gauss' Law
  • Familiarity with electric field calculations for cylindrical geometries
  • Knowledge of surface and volume charge densities
  • Proficiency in using the permittivity of free space (epsilon_0)
NEXT STEPS
  • Study the application of Gauss' Law in different geometries
  • Learn about electric field calculations for non-uniform charge distributions
  • Explore the concept of electric flux and its relation to electric fields
  • Investigate the effects of charge distribution on electric field strength
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or physics research focusing on electric fields and charge distributions.

Soaring Crane
Messages
461
Reaction score
0

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.
 
Physics news on Phys.org
Soaring Crane said:

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.
 

Similar threads

Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
781
  • · Replies 11 ·
Replies
11
Views
2K
Replies
11
Views
2K
Replies
29
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
10
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K