Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 2.40

Two spherical cavities, of radii $a$ and $b$ , are hollowed out from the interior of a (neutral) conducting sphere of radius $R$ (As sen in Fig. 2.49). At the center of each cavity a point charge is placed, call these charges $q_{a}$ and $q_{b}$ .

(a) Find the surface charge densities $σ_{a}, σ_{b}$ and $σ_{R}$ .

(b) What is the field outside the conductor?

(d) What is the force on $q_{a}$ and $q_{b}$ ?

(e) Which of these answers would change if a third charge, $q_{c}$ , were brought near the conductor (but outside of it)?

(a) The set up is such that we can assume that the induced charges get distributed uniformly across their surface. Let's recall that

Area of a sphere = 4 π r^{2}

Where $r$ is the sphere's radius. Because the induced charge has opposite sign ("The total charge induced on the cavity wall is equal and opposite to the charge inside" Page 97), it follows that

\begin{aligned} σ_{a} & = - \frac{q_{a}}{4 π a^{2}} \\ σ_{b} & = - \frac{q_{b}}{4 π b^{2}} \\ σ_{R} & = \frac{q_{a} + q_{b}}{4 π R^{2}} \end{aligned}

(b) Recall that a conductor conceals from us the information concerning the nature of the cavity(ies) and the internal configuration(s), we only get to know the total charge it contains. Since the body we are studying is a sphere, it follows that we can use the formula for the electric field of a uniformly charged sphere (Why uniformly charged? Because the conductor conceals the internal configuration, so it is equally good to consider it to be uniformly charged).

The electric field of a uniformly charged sphere is given by: (If you don't know where this comes from, read Example 2.3, page 69.)

E_{unif. sphere} = \frac{1}{4 π ϵ_{0}} \frac{Q}{r^{2}} \hat{r}

Where $Q$ is the total charge inside. In this case:

Q = q_{a} + q_{b}

Thus

E_{sphere with cavities and charges} = \frac{1}{4 π ϵ_{0}} \frac{q_{a} + q_{b}}{r^{2}} \hat{r}

\begin{aligned} E_{a} & = \frac{1}{4 π ϵ_{0}} \frac{q_{a}}{r^{2}} \hat{r} \\ E_{b} & = \frac{1}{4 π ϵ_{0}} \frac{q_{b}}{r^{2}} \hat{r} \end{aligned}

Where we measure $r$ starting from the charge itself. if this doesn't convince you, consider a (spherical) Gaussian surface around each charge. It is clear that we only have to consider the contributions from the point charge as long as we reside inside the cavity.

(d) Going back to the answer from part (c), we see that to charge $a$ , the $E_{b}$ doesn't even "exist" (the same goes for $b$ not detecting any $E_{a}$ field). As such, it is clear that

F_{a b} = F_{b a} = 0

(e) Let's go one by one:

In part (a), $σ_{a}$ and $σ_{b}$ only depend on their respective charges, so they won't change. $σ_{R}$ will change, as it is a surface that's "talking" to the "cavity" where the new charge resides.

In part (b), $E_{outside}$ will change as we have a new charge.

In part (c), $E_{a}$ and $E_{b}$ are unaffected by the external charge.

In part (d), again taking from part (c), the force between them will still be zero.

Fernando Garcia

Home Research Blog Other About me

PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 2.40

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