The charge on capacitor of capacitance 20 μF as shown in figure given below is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 100 μC
As capacitor are connected in series -
$\frac{1}{C_{net}}=\frac{1}{20}+\frac{1}{10}=\frac{1+2}{20}$
$C_{net}=\frac{20}{3}μF$
Now,
$C=\frac{Q}{V}$
$⇒Q=CV$
$=\frac{20}{3}×15=100 μC$