Answer
(a) $5.0\times 10^{-6}F$
(b) equal charge
(c) $75.0\times 10^{-6}C$
Work Step by Step
(a) We can find the required equivalent capacitance as follows:
$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$
We plug in the known values to obtain:
$\frac{1}{C_{eq}}=\frac{1}{7.5\times 10^{-6}F}+\frac{1}{15\times 10^{-6}F}$
$\implies\frac{1}{C_{eq}}=0.2\times 10^{6}$
$\implies C_{eq}=\frac{1}{0.2\times 10^6}$
$C_{eq}=5.0\times 10^{-6}F$
(b) We know that both capacitors are connected in series; hence, there will be the same amount of charge stored on both capacitors as the net charge of the battery is not distributed.
(c) We know that
$Q=C_{eq}V$
We plug in the known values to obtain:
$Q=5.0\times 10^{-6F}\times 15.0V$
$Q=75.0\times 10^{-6}C$
