Spheres, Plates, and Gauss' Law

In summary: So you don't need to divide by two.In summary, the first problem involves finding the electric field inside a sphere of a given radius and volume charge density. The solution uses the equation E = (k*Q*r)/(R^3), where R is the sphere's radius and r is the distance inside the sphere. The second problem involves finding the electric field 10 µm from a large, flat conducting plate with a given surface charge density. The solution uses the equation E = sigma/epsilon_0, where sigma is the surface charge density and epsilon_0 is the permittivity of free space.
  • #1
Soaring Crane
469
0

Homework Statement



A sphere of radius 8.0 cm carries a uniform volume charge density rho = 500*10^-9 C/m^3. What is the electric field at r = 3.0 cm?

a.36.0 N/C

b.230 N/C

c.140 N/C

d.565 N/C

e.450 N/C




Homework Equations



E = (k*Q*r)/(R^3), where R = sphere’s radius and r is distance inside sphere

The Attempt at a Solution


V = (4/3)*pi*(0.08 m)^3 = 0.00214 m^3
Q = p*V = (500*10^-9 C/m^3)*(0.00214 m^3) = 1.07*10^-9 C

E = (k*Q*r)/(R^3), where R = sphere’s radius and r is distance inside sphere

E = (8.988*10^9)*(1.07*10^-9 C)/(0.03 m^2) = 564.7 N/C




Homework Statement



A large, flat conducting plate has a surface charge density sigma = 8.0*10^-9 C/m^2 on one of its surfaces. What is the magnitude of the electric field 10 µm from this plate?


a.72 N/C

b.0.23 kN/C

c.0.90 kN/C

d.90 MN/C

e.9.0 1012 N/C



Homework Equations



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

The Attempt at a Solution



E*A_plate = q/epsilon_0
E*A_plate = (sigma*A_plate)/(epsilon_0)

E = sigma/epsilon_0 = (8.0*10^-9 C/m^2)/(8.85*10^-12) = 904 N/C

Thanks.
 
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  • #2
This isn't really a question. Is it?
 
  • #3

What is a sphere?

A sphere is a three-dimensional shape that is perfectly round and has all points on its surface equidistant from its center. It can be thought of as a collection of points that are all the same distance away from a central point.

What is a plate?

A plate is a two-dimensional object that has a constant thickness and can be thought of as a flat, infinite surface. It is often used in physics and mathematics to represent an idealized surface with uniform properties.

What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It states that the electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.

How does Gauss' Law relate to spheres and plates?

Gauss' Law can be used to calculate the electric field at any point outside of a charged spherical or planar surface, as long as the charge distribution is symmetric. This is because the electric field is constant in magnitude and direction at all points on the surface, making it easier to calculate using Gauss' Law.

What are the applications of spheres, plates, and Gauss' Law in science and engineering?

Spheres, plates, and Gauss' Law have a wide range of applications in various fields, including electromagnetism, electrostatics, and fluid mechanics. They are used to model and analyze systems with symmetrical charge distributions, such as electric circuits, capacitors, and conductors. They also have applications in optics, where they can be used to calculate the electric field of a spherical or planar lens. In engineering, Gauss' Law is used in the design of electronic devices and power systems.

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