Question

Problem 4. A solid conductor has three spherical cavities. Each cavity contains a point charge as shown The charge on the outer surface of the conductor is 12.0 nC 12 nC 20 nC 15 n -30 n (a) What is the charge on the surface of each cavity? (b) What is the excess (added) charge on the conductor? (c) What is the field at a point in the cavity on the left located 2.00 cm from the the 20nC charge? (Note that the point is located in the cavity.)

Answer 1

a)

Let us consider one of the cavities.

construct a Gaussian surface around the cavity.

and apply the integral form of Gauss Law

$\oint \vec E.d\vec a=q_{net-enclosed}$

But this Gaussian surface lies inside the conductor, and inside the conductor, the field is 0, hence E is 0 over the Gaussian surface.

$0=q_{net-enclosed}$

but the net charge enclosed is

$q+q_{induced}=0\\ q_{induced}=-q$

where q is the charge inside the cavity and q_induced is the charge induced on the surface of the cavity.

Hence the charges on the cavities are equal and opposite to the charge inside the respective cavity.

The charges induce are -20nC, +30nC, -15nC

b)

corresponding to each cavity, a net equal and opposite charge would be induced on the outer surfa+ce.

Hence net charge induce on the outer surface is

+20nC-30nC+15nC= +5nC.

Hence the excess charge (added ) is ZERO, as corresponding to charge induced on each cavity, there is equal and opposite charge induced on the surface also.

_{[shown in the above diagram in blue colour]}

c)

The cavity shields the effect of the charges OUT of the cavity, and also the charge induces inside the cavity arrange themselves in such a manner that the electric field due to these charges is ZERO inside the cavity. Hence the only charge contributing to the electric field is the charge inside the cavity.

Hence, Using The formula for the electric field for a point charge we can easily find the field inside the cavity.

$\vec E=\frac{1}{4\pi \epsilon_0}\times \frac {q}{r^2}\hat r$

where q is the charge inside the cavity and $\vec r$ is the vector joining the point to the location of charge q.

[note :this is valid as long as $\vec r$ is inside the cavity]

$\\ \vec E=\frac{1}{4\pi \epsilon_0}\times \frac {20\times 10^{-9}C}{(2\times 10^{-2}m)^2}\hat r \\ \\ \vec E=449.377 \ kV/m$