Spheres, Plates, and Gauss' Law

AI Thread Summary
The discussion addresses two physics problems involving electric fields generated by a charged sphere and a conducting plate. For the sphere with a radius of 8.0 cm and a charge density of 500*10^-9 C/m^3, the electric field at 3.0 cm was calculated to be approximately 565 N/C. In the second problem, a flat conducting plate with a surface charge density of 8.0*10^-9 C/m^2 produced an electric field of about 904 N/C at a distance of 10 µm. The calculations were confirmed as correct, and additional resources were provided for further reference. The discussion emphasizes the application of Gauss' Law in determining electric fields in both scenarios.
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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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