$6\mathrm{.}25($a$)$ Show that the voltage$-$division rule for two capacitors in series as in Fig. $6\mathrm{.}59($a$)$ is

$6\mathrm{.}65$

$v_1=\frac{C_2}{C_1+C_2}{v}_{sv},v_2=\frac{C_1}{C_1+C_2}{v}_s$

assuming that the initial conditions are zero.

$($a$)$

Figure $6\mathrm{.}59$

For Prob. $6\mathrm{.}25.$

$($b$)$ For two capacitors in parallel as in Fig. $6\mathrm{.}59($b$),$ show that the current$-$division rule is

$i_1=\frac{C_1}{C_1+C_2}{i}_s,i_2=\frac{C_2}{C_1+C_2}{i}_s$

assuming that the initial conditions are zero.