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cyberstudent
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A metal sphere of radius R with a charge q is surrounded by a concentric metal sphere as shown in Figure P.23. The outer surface of the spherical shell has a charge of + 30.0 μC and the inner surface of the shell has a charge of +25.0 μC.
b is the radius of the inner surface of the shell and a is the radius of the outer surface of the shell.
(a) Find q
q = -25.0 μC, since a conductor has no electric field. Only the charge on the inner surface of the outer shell affects the metal sphere inside.
(b) Sketch qualitative graphs of (i) the radial electric field component Er and (ii) the electric potential V as functions of r.
This is trickier. Let’s start with V(R)
For r < R (inside sphere), V(r) = -25.0 μC/(k R)
This is because the interior of a conductor has the same potential everywhere. The potential is equal to the potential on the surface.
For R < r < b, the potential goes from -25.0 μC/(k R) to + -25.0 μC/(k R). Right? Very unsure about this.
For b < r < a, the potential difference is constant. V(r) = 30.0 μC - 25.0 μC = 5 μC.
Hmmm?
For a < r, the potential difference follows a curve from 30 μC/(k R) to 0 as r nears infinity.
Right?
OK. I know I messed up but I’m still going to try E(R)
For r < R (inside sphere), E(r) = 0
The electric field within a conductor is always 0.
For R < r < b, the potential goes from -25.0 μC/(k R^2) to + -25.0 μC/(k R^2). Right? Very unsure about this.
For b < r < a, E(r) = 0
The electric field within a conductor is always 0.
For a < r, the potential difference follows a curve from 30 μC/(k R^2) to 0 as r nears infinity.
I uploaded my graphs, but I’m not sure if people are able to see my attachments.
b is the radius of the inner surface of the shell and a is the radius of the outer surface of the shell.
(a) Find q
q = -25.0 μC, since a conductor has no electric field. Only the charge on the inner surface of the outer shell affects the metal sphere inside.
(b) Sketch qualitative graphs of (i) the radial electric field component Er and (ii) the electric potential V as functions of r.
This is trickier. Let’s start with V(R)
For r < R (inside sphere), V(r) = -25.0 μC/(k R)
This is because the interior of a conductor has the same potential everywhere. The potential is equal to the potential on the surface.
For R < r < b, the potential goes from -25.0 μC/(k R) to + -25.0 μC/(k R). Right? Very unsure about this.
For b < r < a, the potential difference is constant. V(r) = 30.0 μC - 25.0 μC = 5 μC.
Hmmm?
For a < r, the potential difference follows a curve from 30 μC/(k R) to 0 as r nears infinity.
Right?
OK. I know I messed up but I’m still going to try E(R)
For r < R (inside sphere), E(r) = 0
The electric field within a conductor is always 0.
For R < r < b, the potential goes from -25.0 μC/(k R^2) to + -25.0 μC/(k R^2). Right? Very unsure about this.
For b < r < a, E(r) = 0
The electric field within a conductor is always 0.
For a < r, the potential difference follows a curve from 30 μC/(k R^2) to 0 as r nears infinity.
I uploaded my graphs, but I’m not sure if people are able to see my attachments.