A -3.30-μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius of 2.50 cm and an outer radius of 3.50 cm. The shell has a net charge of -5.30 μC.
1) Determine the charge on inner surface of shell. (Express your answer to three significant figures.)
2) Determine the charge on outer surface of shell. (Express your answer to three significant figures.)
3) Determine the electric field just outside the shell. (Express your answer to three significant figures.)

1) +3.30 μC    2) -8.60 μC   3) 6.31*10⁷ N/C

Explanation:

1) Applying Gauss´Law to a gaussian spherical surface with a radius between the inner radius of the shell, and the outer radius, we find that the net flux through this surface is just zero:

This is due to no electric field can exist inside a conductor in electrostatic condition.

So, if the left side of the equation is 0, the right side must be zero too:

If the total charge enclosed by the gaussian surface is 0, and we have a charge of -3.30 μC in the center, there must exist a +3.30 μC, on the inner surface of the shell, so the total charge be 0.

⇒ Qin = +3.30 μC

2) As the charge must be conserved, if the net charge of the shell is -5.30 μC, if we have +3.30 μC on the inner surface, we will have a charge on the outer surface, that keep the total charge of the shell constant, as follows:

Qou = Qsh - Qin = -5.30 μC -(+3.30 μC) = -8.60 μC

3) In order to find the electric field just outside the shell, we can apply Gauss' Law to a gaussian spherical surface of radius equal to the outer  shell radius.

By symmetry, as the electric field must be radial, at a same radius, the magnitude of the electric field must be the same. As the dA vector is normal to the surface, is also radial, so we can write Gauss´ Law as follows:

E* 4*π*r² = Qenc / ε₀

where Qenc = Qc + Qin + Qou

As Qc+ Qin = 0, ⇒ Qenc =Qou = -8.60 μC

r = distance to the center of the sphere = 3.5 cm = 0.035 m

ε₀ = 8.85*10⁻¹² C²/N*m²

Replacing by the values, we  can solve for E, as follows:

⇒ E = 6.31*10⁷ N/C

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )

it is called kinetic energy