Context: this is a multiple choice question, but no explanation is given

Question: three capacitors of equal capacitance are connected in the following circuit. Determine the ratio of charge in capacitor B with respect to capacitor D. enter image description here

According to the answers, the charge in B is half that stored in D. Intuitively, this makes sense (i.e. Kirchoff's Law), but is it possible/necessary to explain the problem using capacitance equations (i.e. $C=\frac{q}{V}$)? I'd just like to know out of understanding's sake. Thanks


1 Answer 1


The two capacitors in parallel acts as if you only got one $C=C_B+C_F=2C_B$ which is in series with the third $C_D=C_B$.

Then since you have a series the same amount of charge is stored in each of the series*: $Q_{BF}=Q_D$. But since $Q_{BF}=2Q_B$ you have that $Q_B=\frac{Q_D}{2}$.

*(since charge is proportional to current in a length of time the capacitance itself plays no role in determining charge unless it was previously charged -which happens rarely in homeworks but quite seldomly in real life-, that is why a series of capacitances store the same amount of Q on each plate regardeless to the value of C)

  • $\begingroup$ Thanks for the simple and concise explanation. Many thanks, it makes sense to me now :) $\endgroup$
    – tim9800
    Jan 7, 2019 at 9:33
  • $\begingroup$ You welcome! remember: elements in parallel share the same voltage across them, elements in series share the same current flowing in each one, same here! $\endgroup$ Jan 7, 2019 at 9:38

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