In the figure a solid sphere of radius a = 3.00 cm is concentric with a spherica
ID: 2129444 • Letter: I
Question
In the figure a solid sphere of radius a = 3.00 cm is concentric with a spherical conducting shell of inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net uniform charge q1 = +7.17 fC; the shell has a net charge q2 =
In the figure a solid sphere of radius a = 3.00 cm is concentric with a spherical conducting shell of inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net uniform charge q1 = +7.17 fC; the shell has a net charge q2 = -q1. What is the magnitude (in N/C) of the electric field at radial distances (a)r = 0 cm, (b)r = a/2.00, (c)r = a, (d)r = 1.50a, (e)r = 2.30a, and (f)r = 3.50a? What is the net charge (in C) on the (g) inner and (h)outer surface of the shell? The answer for part a, part e, part f and part h are all zero. I have confirmed this with my teacher. I only need help with parts B, C and D. The answer for part g is -7.17E-15Explanation / 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
Surface area of a sphere: A = 4*pi*r^2
Volume of a sphere: V = (4/3)*pi*r^3
Find the charge density of the solid sphere. Let charge density of the sphere be 'P'. Let the radius of the solid sphere be 'a'. 'Q' is the charge of object. 'V' is the volume of the object.
a = r1 = 2.00 x 10^-2 m
P = Q / V
= Q / (4/3)*pi*a^3
= (10.30 x 10^-6 C) / (4/3)*(3.14)*(2.00 x 10^-2 m)^3
= 0.3074 C/m^3
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a) Remember that there is no charge inside of a solid conductor. Therefore...
|E| = 0
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b) 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| = (10.30 x 10^-6 C) / 4*(3.14)*(8.85 x 10^-12 C^2/N*m^2)*(3.00 x 10^-2 m)^2
= 1.03 x 10^16 N/C
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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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d) We already know we treat the electric outside of a sphere like a point charge when solving for electric field. This time, we have 2 different charges coming from 2 concentric spheres. To get the electric field, we take the sum of the two electric fields from the radius indicated in the problem.
r = 7.00 x 10^-2 m , radius of the Gaussian surface
Q_inner = q1 = 10.30 x 10^-6 C
Q_outer = q2 = -4.00 x 10^-6 C
|E_inner| = Q_inner / 4*pi*epsilon_0*r^2
|E_outer| = Q_outer / 4*pi*epsilon_0*r^2
|E_net| = |E_inner| + |E_outer|
= [Q_inner / 4*pi*epsilon_0*r^2] + [Q_outer / 4*pi*epsilon_0*r^2]
= [Q_inner+Q_outer] / [4*pi*epsilon_0*r^2]
= [(10.30 x 10^-6 C)-(4.00 x 10^-6 C)] / [4*(3.14)*(8.85 x 10^-12 C^2/N*m^2)*(7.00 x 10^-2 m)^2]
= 1.12 x 10^7 N/C
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