The next TWO question apply to the following: A solid conducting sphere of radiu
ID: 1586709 • Letter: T
Question
The next TWO question apply to the following: A solid conducting sphere of radius 2.0 cm has a charge of +4 HC. A conducting spherical shell of inner radius 4.0 cm and outer radius 5.0 cm is concentric with the solid sphere and has a charge of -3.0 HC. 6. Find the electric field at a radius of 3.0 cm from the center of this charge configuration. A 9.99e +006 N/C B 2.92 +008 N/C C 4.00e+007 N/C 3.60e +008 N/C E 0.0 N/C 7. Find the electric field at a radius of 4.5 cm from the center of this charge configuration. A 0.0 N/C -2.47e+006 N/C C 4.44 006 N/C D 2.25 +007 N/C E 1.78e +007 N/CExplanation / Answer
r1 = 2.00 x 10^-2 m , radius of inner solid sphere
q1 = 10.30 x 10^-6 C , charge on inner solid sphere
r2 = 4.00 x 10^-2 m , inner radius of shell
r3 = 5.00 x 10^-2 m , outer radius of shell
q2 = -4.00 x 10^-6 C , charge on shell
a) We can use Gauss' law, but know that when the distance away from a charged sphere is larger than its radius (assuming the sphere is the only object in space), the sphere can be treated as a point charge. 'a' is the radius of the sphere. 'r' is the radius of the Gaussian surface. '|E|' is the magnitude of the electric field, 'A' is the surface area of the Gaussian surface, 'Q_enc' is the charge enclosed, and 'eplison_0' is the vacuum permittivity constant.
r = 3.00 x 10^-2 m , radius of the Gaussian surface
|E|*A = Q_enc / epsilon_0
|E|*4*pi*r^2 = (4*P*pi*a^3) / 3*epsilon_0
|E|*r^2 = (P*a^3) / 3*epsilon_0
|E| = (P*a^3) / 3*r^2*epsilon_0
substitute P...
|E| = [(Q / (4/3)*pi*a^3)*a^3] / 3*r^2*epsilon_0
It's confusing to look at since it's all text, but the a^3 cancels out and the 3's cancel out. We are left with...
|E| = Q / 4*pi*epsilon_0*r^2
As you can see, it's the same equation for an electric field from a point charge (as stated earlier), so we will use this as a reference for the rest of the problem.
|E| = (4 x 10^-6 C) / 4*(3.14)*(8.85 x 10^-12 C^2/N*m^2)*(3.00 x 10^-2 m)^2
= 3.99 x 10^7 N/C is equal to 4*10^7 N/C
B) c) When r = 4.50 cm, this means we draw a Gaussian surface inside of the thickness of the spherical shell. Remember the properties of conductors. There is no charge inside the solid of a conductor, so there is no electric field.
|E| = 0
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