Irodov 3.65: Zero Potential, Yet Still Charged – Amit & Deepali Sinha Foundation

Problem Irodov 3.65

Two conducting spheres share a center. The inner one, of radius $a$ , holds a positive charge $q_1$ . Find the charge $q_2$ that must be placed on the outer sphere, of radius $b$ , to bring the potential of the inner sphere down to zero. For that case, find how the potential $\varphi$ varies with the distance $r$ from the center, and sketch the graph of $\varphi(r)$ .

The two concentric conducting spheres: the inner one, radius a, carrying q_1, and the outer one of radius b. — The two concentric conducting spheres: the inner one, radius $a$ , carrying $q_1$ , and the outer one of radius $b$ .

Given

inner sphere: radius $a$ , charge $q_1 > 0$
outer sphere: radius $b$

Find

the charge

q_2

that makes the inner sphere’s potential zero, and

\varphi(r)

with its graph

Solution

Step 1The potential of a uniformly charged sphere

By symmetry each sphere’s charge spreads uniformly over it. A uniform spherical charge $Q$ of radius $R$ acts outside as a point charge at the center, and inside it the field is zero, so the potential is constant there at its surface value.

\varphi_Q(r) = \frac{Q}{4\pi\varepsilon_0 R} \quad (r \le R), \qquad \varphi_Q(r) = \frac{Q}{4\pi\varepsilon_0 r} \quad (r \ge R) \tag{1}

Step 2The charge that zeroes the inner sphere’s potential

On the inner sphere, $r = a$ , the charge $q_1$ contributes its surface value and $q_2$ , whose radius $b$ is larger, its constant interior value. By (1) the sum is set to zero:

\begin{aligned} \varphi(a) &= \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{a} + \frac{q_2}{b}\right) = 0\\ \frac{q_2}{b} &= -\frac{q_1}{a} \end{aligned}

q_2 = -\dfrac{b}{a}\,q_1

(2)

The outer sphere must carry a negative charge larger in magnitude than $q_1$ , since $b > a$ .

Step 3The potential in each region

Superpose (1) for the two spheres, region by region. Inside the inner sphere the potential equals its surface value, zero. Between the spheres, $q_1$ acts as a point charge and $q_2$ gives its constant:

\varphi(r) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r} + \frac{q_2}{b}\right) \qquad (a \le r \le b)

Outside both, the total charge acts as a point charge at the center:

\varphi(r) = \frac{q_1 + q_2}{4\pi\varepsilon_0 r} \qquad (r \ge b)

Substituting $q_2$ from (2) into both and collecting the three regions:

\varphi(r) = \dfrac{q_1}{4\pi\varepsilon_0}\begin{cases} 0, & r \le a\\[2pt] \dfrac{1}{r} - \dfrac{1}{a}, & a \le r \le b\\[6pt] \dfrac{1 - b/a}{r}, & r \ge b \end{cases}

(3)

Step 4The graph

φ(r), drawn for b = 2a: zero up to r = a, falling as 1/r - 1/a to its lowest value at r = b, then rising back toward zero as -1/r. — $\varphi(r)$ , drawn for $b = 2a$ : zero up to $r = a$ , falling as $1/r - 1/a$ to its lowest value at $r = b$ , then rising back toward zero as $-1/r$ .

The potential is zero up to $a$ , falls between the spheres, and is lowest on the outer sphere, where both pieces of (3) give the same value:

\varphi(b) = \frac{q_1}{4\pi\varepsilon_0}\left(\frac{1}{b} - \frac{1}{a}\right) = \frac{q_1}{4\pi\varepsilon_0}\,\frac{1 - b/a}{b} < 0

Beyond $b$ it climbs back toward zero as $1/r$ ; the slope breaks at $a$ and at $b$ , where the charged surfaces sit.

Check

The field from (3) must be that of the charge enclosed. Between the spheres,

E = -\frac{d\varphi}{dr} = \frac{q_1}{4\pi\varepsilon_0 r^2},

the field of $q_1$ alone, pointing outward. Outside,

E = -\frac{d\varphi}{dr} = \frac{q_1}{4\pi\varepsilon_0}\,\frac{1 - b/a}{r^2} < 0,

pointing inward, the field of the negative total $q_1 + q_2$ . A form growing in proportion to $r$ beyond $b$ would instead never vanish at infinity, where every potential here is referred to zero.

Remark

The inner sphere sits at zero potential and still carries its full charge $q_1$ : zero potential means the same potential as infinity, not zero charge. In fact the situation is exactly that of an earthed inner sphere: connect it to infinity through a thin wire, put $q_2$ on the outer sphere, and the inner one draws up the charge $q_1 = -(a/b)\,q_2$ , the same relation read the other way.