Question

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density is?

Answer 1

`1.57 * 10^(3) Q`

Step-by-step explanation:

The volume charge density is defined by ρ =
`(Q)/(V)` (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are
`(Coulombs)/(m^(3) )`, so we have to express the radius in meters:

inner radius =
`4 cm * (1 m)/(100 cm) = 0.04m`

outer radius =
`6 cm * (1m)/(100cm)  = 0.06m`

Now, we know that the volume of the sphere is calculated by the formula:

`V = (4)/(3)\pi r^(3)`, and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V =
`(4)/(3)\pi (r_(o) ^(3) - r_(i) ^(3))`, where
`r_(o)` is the outer radius and
`r_(i)` is the inner radius.

Replacing the volume formula in the Equation A:

ρ =
`(Q)/((4)/(3)\pi(r^(3) _(o)-r_(i) ^(3)))`

ρ =
`(3Q)/(4\pi (r_(o) ^(3)-r_(i) ^(3) ) )`

Replacing the values of the outer and inner radius whe have:

ρ =
`(3Q)/(4\pi (1.52 * 10^(-4)))`

ρ =
`1.57 * 10^(3) Q`