Gauss's Law

1. Answer it.
2. Answer it.
3. Answer it.

1. The charge will distribute itself so as to "screen out" the electric field within the hollow sphere. There will be just as much charge on the inside of the surface as there is on the nested sphere with the total charge adding up to −6 μC. Thus there must be…

   1. −12 μC of charge on the inner surface and…

   2. +6 μC of charge on the outer surface.

2. Now that we know the charge distribution we can determine the electric field by repeated application of Gauss' law. Imagine a spherical Gaussian surface concentric with the nested spheres and watch its radius vary from 0 to 10 cm.

   ∯ E · dA =	Q
   	ε0
   E (4πr2) =	Q
   	ε0

   E =	1		Q
   	4πε0		r2
   E =	kQ
   	r2

   For the first 3 cm the Gaussian sphere contains no charge, which means there is no electric field. That's the way it works in a conductor. There can be no field inside a conductor once the charges find their equilibrium distribution.

   E =  kQ r2

   E =  (9.0 × 109 N m2/C2)(0 C) r2

   E = 0 V/m, 0 cm to 3 cm

   When the radius reaches 3 cm the Gaussian sphere finally contains some charge.

   Q_net = +12 μC

   No more charge is added to the Gaussian sphere until the radius reaches 6 cm.

   E =  kQ r2

   E =  (9.0 × 109 N m2/C2)(+12 × 10−6 C) r2

   E =  +108,000 V m , 3 cm to 6 cm r[m]2

   When the radius reaches 6 cm, the Gaussian surface contains no net charge.

   Q_net = (+12 μC) + (−12 μC) = 0 μC

   No charge, no field. Or is it the other way round? No field inside a conductor, therefore no net charge inside the Gaussian surface.

   E =  kQ r2

   E =  (9.0 × 109 N m2/C2)(0 C) r2

   E = 0 V/m, 6 cm to 8 cm

   When the Gaussian surface makes it to the outside of the outer sphere, the net charge is nonzero once again. This time it's…

   Q_net = (+12 μC) + (−6 μC) = +6 μC

   Thus…

   E =  kQ r2

   E =  (9.0 × 109 N m2/C2)(+6 × 10−6 C) r2

   E =  +54,000 V m , 8 cm to ∞ r[m]2

   Numbers in. Answers out. See the table below for the summary of all this talk. By the way, the positive signs in the answers tell us that the field is directed radially outward in the places where it exists.

   Now, on to the potential. In this problem, it's best to compute electric potential from the electric field using the mutated form of the work-energy theorem. Imagine a small, positive test charge. Drag it from infinity to any point a distance r from the center of the spheres. The work per unit charge is the electric potential and the force per unit charge is the field.

   	r				r
   V = −	⌠ ⎮ ⌡	E ·dr	⇐	U = −	⌠ ⎮ ⌡	F · dr
   	∞				∞

   Apply this formula in a piecewise manner, noting the discontinuities in the electric field at the surfaces of the conductors. Work from the "outside in" as is the custom with potentials.

   For the space surrounding the outer sphere we get…

   r V = −  ⌠ ⎮ ⌡ kQ  dr r2 ∞

   V = −  (9.0 × 109 N m2/C2)(+6 × 10−6 C) r[m]

   V =  +54,000 V m , ∞ to 8 cm r[m]

   Electric potential in a conductor is the same everywhere, is the same as the potential at the last place it was computable — the surface.

   0.08 m V = −  ⌠ ⎮ ⌡ kQ  dr r2 ∞

   V = −  (9.0 × 109 N m2/C2)(+6 × 10−6 C) 0.08 m

   V = 0.675 MV, 6 cm to 8 cm

   Now, in the gap between the conductors we see the first application of piecewise integration.

   0.08 m 0.06 m r V = −  ⌠ ⎮ ⌡ E · dr  −  ⌠ ⎮ ⌡ E · dr  −  ⌠ ⎮ ⌡ E · dr ∞ 0.08 m 0.06 m

   V = + 0.675 MV + 0 MV + kQ ⎡ ⎢ ⎣ 1  −  1 ⎤ ⎥ ⎦ r[m] 0.06 m

   V = + 0.675 MV + 0.108 MV m ⎡ ⎢ ⎣ 1  −  1 ⎤ ⎥ ⎦ , 3 cm to 6 cm r[m] 0.06 m

   My, that was unpleasant.

   We finish by evaluating this last expression at 3 cm. That's the potential from the surface of the inner sphere all the way to the center.

   V = + 0.675 MV + 0.108 MV m ⎡ ⎢ ⎣ 1  −  1 ⎤ ⎥ ⎦ 0.03 m 0.06 m

   V = 2.475 MV, 0 cm to 3 cm

   Check the table below for the results of all this work.

   distance from center (cm)	electric field (MV/m)		electric potential (MV)
   	direction	magnitude
   0	n/a	0	+2.475
   1	n/a	0	+2.475
   2	n/a	0	+2.475
   3	n/a, out	0, 120	+2.475
   4	out	67.5	+1.575
   5	out	43.2	+1.035
   6	out, n/a	30.0, 0	+0.675
   7	n/a	0	+0.675
   8	n/a, out	0, 8.44	+0.675
   9	out	6.67	+0.600
   10	out	5.40	+0.540

3. Use this information to make graphs of the…

   1. magnitude of the electric field

   2. electric potential

Answer it.

Answer it.