Question

25 capacitors each of capacitance \(1\ \mu\text{F}\) are connected in series to a battery of \(100\ \text{V}\). The total charge stored on the capacitors is

• A. \(2.0\times10^{-5}\ \text{C}\)
• B. \(2.5\times10^{-3}\ \text{C}\)
• C. \(4.0\times10^{-6}\ \text{C}\)
• D. \(1.5\times10^{-6}\ \text{C}\)

Hint:

For \(n\) identical capacitors in series, \[ C_{\text{eq}}=\frac{C}{n}. \] The charge on every capacitor in a series combination is the same and is given by \[ Q=C_{\text{eq}}V. \]

Answer 1

The Correct Option is C

Step 1: Find equivalent capacitance for series combination.
25 capacitors of \( 1\,\mu\text{F} \) in series: \( C_{eq} = \frac{1}{25}\,\mu\text{F} = 0.04\,\mu\text{F} \).

Step 2: Find total charge.
\( Q = C_{eq}\times V = 0.04\times10^{-6}\times100 = 4\times10^{-6}\text{ C} \)

\[ \boxed{Q = 4\times10^{-6}\text{ C}} \]