# Effect on capacitance when capacitor number increased [closed]

What happens to total charge and potential difference when number of capacitors increased in series and parallel combination? My question in text book is:

Initially, a single capacitance $C_1$ is wired to a battery. Then capacitance $C_2$ is added in parallel. Are

a:the potential difference across $C_1$ and

b:the charge $q_1$ in $C_1$ now more than,less than,or the same as previously?

c:is the equivalent capacitance $C_{12}$ of $C_1$ and $C_2$ more than,less than or equal to $C_1$?

d:Is the charge stoed on $C_1$ and $C_2$ together more than,less than,or equal to the charge stored previously on $C_1$?

Repeat this for series combination?

## closed as off-topic by Martin, ACuriousMind♦, Kyle Kanos, Qmechanic♦May 30 '15 at 19:28

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• You're question seems a little vague and incomplete. You have not specified whether the potential difference is kept constant(capacitors are connected across a battery) or charge is kept constant(they are isolated).Please mention a specific situation along with initial conditions and only then someone would be really able to assist you. – user2369284 May 30 '15 at 17:39
• Okay..thankyou for mentioning..i will rewrite my question – Sadia Tahir May 30 '15 at 18:11

Consider capacitors in series with a voltage source. Kirchhoff's voltage law tells you that the sum of voltage drops and rises on the capacitors should equal the sum of voltage of the source. This holds for any number of capacitors, thus if you increase the number of capacitors, nothing will happen to total voltage difference. The net capacitance of $n$ capacitors in series are found by: \begin{equation}\frac{1}{C_\sum}=\frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}=\sum_{i=1}^n\frac{1}{C_i}\end{equation} \begin{equation}C_\sum = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}}\end{equation}

From this you may see that if you add one more capactior with capacitance $C_{n+1}$, the net capacitance will change: \begin{equation}C'_\sum = \frac{1}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} + \frac{1}{C_{n+1}}}\end{equation} The total charges in the two cases: \begin{equation}Q= V*C_\sum = \frac{V}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}}\end{equation} \begin{equation}Q' = V*C'_\sum = \frac{V}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} + \frac{1}{C_{n+1}}}\end{equation} The change in total charge is: \begin{equation}\Delta Q = Q'-Q = V*C'_\sum - V*C_\sum= \frac{V}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}} - \frac{V}{\frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} + \frac{1}{C_{n+1}}}\end{equation}

Consider now the cases of capacitors in parallel and in parallel with a voltage source. This is much simpler, since the net capacitance of capactiors in parallel are just the sum of capacitances: \begin{equation}C_\sum=C_1+C_2+\ldots+C_n\end{equation} If you add one more, it will change to \begin{equation}C'_\sum=C_1+C_2+\ldots+C_n+C_{n+1}\end{equation} Again, because of Kirchhoff's voltage law, the total voltage drop on the capacinates will not change. Thus the change in charge here is: \begin{equation}\Delta Q = Q'-Q = V*C'_\sum-V*C_\sum=V*(C'_\sum-C_\sum)=V*\left[C_1+C_2+\ldots+C_n+C_{n+1} - (C_1+C_2+\ldots+C_n)\right]=V*C_{n+1}\end{equation}

Basically, the resultant capacitance is give by (you probably already know this ) :
1) n capacitors in series : $1/C_{eq} = 1/C_1 + 1/C_2 + ... + 1/C_n$
clearly $C_{eq}$ decreases and $Q$ decreases.
2) n capacitors in parallel : $C_{eq} = C_1 + C_2 + C_3 + ... C_n$ and hence $Q$ increases.

In case 2 $Q$ is sum of charges stored in all capacitors, but charge on ith capacitor is given by $Q_i = C_iV$

I'm not totally solving it but just providing you with the formulas.

For any capacitor $q = CV$ where V is potential difference across the plates of capacitor.

In series combination of capacitors, charge is same on every capacitor.$C_{eq} = \frac{1}{\sum{\frac{1}{C_i}}}$

In parallel combination of capacitors, potential difference is same across every capacitor. $C_{eq} = \sum{C_i}$

These formulas coupled with the given condition will solve the question.

Hope it helps.