Question

25 capacitors each of capacitance 1 $\mu$F are connected in series to a battery of 100 V. The total charge stored on capacitors is

• A. $2.0 \times 10^{-5}$ C
• B. $2.5 \times 10^{-3}$ C
• C. $4.0 \times 10^{-6}$ C
• D. $1.5 \times 10^{-6}$ C

Hint:

For capacitors in series: $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}$. If all $n$ capacitors are identical ($C$), then $C_{eq} = C/n$.
For capacitors in parallel: $C_{eq} = \sum C_i$.
Charge stored: $Q = CV$.
In a series combination, the charge $Q$ is the same on each capacitor and is equal to the total charge supplied by the source to the combination.
The voltage across each capacitor in series can be different ($V_i = Q/C_i$), and $\sum V_i = V_{total}$.

Answer 1

The Correct Option is C

When capacitors are connected in series, the reciprocal of the equivalent capacitance ($C_{eq}$) is the sum of the reciprocals of individual capacitances ($C_i$): $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}$. In this case, we have $n=25$ capacitors, each with capacitance $C = 1 \text{ } \mu\text{F}$. So, $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \dots \text{ (25 times)} = \frac{25}{C}$. $C_{eq} = \frac{C}{25}$. Given $C = 1 \text{ } \mu\text{F} = 1 \times 10^{-6} \text{ F}$. $C_{eq} = \frac{1 \times 10^{-6} \text{ F}}{25}$. The capacitors are connected to a battery of voltage $V = 100 \text{ V}$. The total charge ($Q_{total}$) stored in the equivalent capacitor is given by $Q_{total} = C_{eq} V$. $Q_{total} = \left(\frac{1 \times 10^{-6}}{25}\right) \text{ F} \times 100 \text{ V}$. $Q_{total} = \frac{100 \times 10^{-6}}{25} \text{ C}$. $Q_{total} = \frac{100}{25} \times 10^{-6} \text{ C} = 4 \times 10^{-6} \text{ C}$. When capacitors are connected in series, the charge stored on each individual capacitor is the same, and this charge is equal to the total charge supplied by the battery to the series combination. So, the charge stored on *each* capacitor is $4 \times 10^{-6}$ C. The question asks for "The total charge stored on capacitors". This can be interpreted in two ways: 1. The sum of charges on all capacitors: This would be $25 \times Q_{each}$ if we consider magnitude on one plate of each. However, for a single capacitor, one plate has $+Q$ and other has $-Q$, net charge is 0. This interpretation is usually not what is meant. 2. The equivalent charge stored by the series combination, which is the charge on each capacitor: $Q_{total}$ as calculated above. This is the standard interpretation for "total charge stored". The phrase "total charge stored on capacitors" in the context of series connection usually means the charge $Q$ that flows from the battery, which then resides on each capacitor (positive plate of first, negative plate of last, and induced charges in between). Thus, $Q = 4 \times 10^{-6} \text{ C}$. \[ \boxed{4.0 \times 10^{-6} \text{ C}} \]