Question

Figure 3.114 shows three concentric spherical conductors A, B, and C with radii R, 2R, and 4R, respectively. A and C are connected by a conducting wire, and B is uniformly charged (charge = +Q). Find

a. charges on conductors A and C, and

b. potentials of A and B

Answer 1

Let the charge on $A$ and $C$ be $q_1$ and $q_2$, respectively (Fig. S3.21). From conservation of charge, we have $q_1+q_2=0$
Hence, $q_1=-q_2$ Since $A$ and $C$ are connected by a conducting wire,so they have the same potential.
$V_A$ = potential of A
$=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{R}+\frac{1}{4\pi {\epsilon }_0}\frac{Q}{2R}+\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{4R}$
$V_C$ = potential of C
$=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{4R}+\frac{1}{4\pi {\epsilon }_0}\frac{Q}{4R}+\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{4R}$
Equalizing the potentials of $A$ and $C$, i.e., $V_A=V_C$ we get
$\frac{q_1}{R}+\frac{Q}{2R}+\frac{q_2}{4R}=\frac{q_1}{4R}+\frac{Q}{4R}+\frac{q_2}{4R}$
or $4q_1+2Q=q_1+Q$
or $q_1=Q/3$
Hence, $q_2=Q/3$.
b. $V_A=\frac{1}{4\pi {\epsilon }_{0}R}[-\frac{Q}{3}+\frac{Q}{2}+\frac{Q}{12}]=\frac{Q}{16\pi {\epsilon }_{0}R}$
$V_B=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{2R}+\frac{1}{4\pi {\epsilon }_0}\frac{Q}{2R}+\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{4R}$
$=\frac{1}{8\pi {\epsilon }_{0}R}[\frac{-Q}{3}+Q+\frac{Q}{6}]=\frac{5Q}{48\pi {\epsilon }_{0}R}$