25 capacitors each of capacitance $1 \mu \mathrm{F}$ are connected in series to a battery of $100…

25 capacitors each of capacitance $1 \mu \mathrm{F}$ are connected in series to a battery of $100 \mathrm{~V}$. The total charge stored on capacitors is
  1. $2.0 \times 10^{-5} \mathrm{C}$
  2. $2.5 \times 10^{-3} \mathrm{C}$
  3. $4.0 \times 10^{-6} \mathrm{C}$
  4. $1.5 \times 10^{-6} \mathrm{C}$

Solution

The given situation is shown below.
In series combination, equivalent capacity is given by $ \begin{aligned} \frac{1}{C_{\mathrm{eq}}} & =\frac{1}{C_1}+\frac{1}{C_2}+\ldots .+\frac{1}{C_{25}} \\ & =\frac{1}{1 \times 10^{-6}}+\frac{1}{1 \times 10^{-6}}+\ldots .+25 \text { terms } \\ & =\frac{25}{1 \times 10^{-6}} \quad\left(\therefore C=1 \times 10^{-6} \mathrm{~F}\right) \end{aligned} $ or $C_{\mathrm{eq}}=\frac{1}{25} \times 10^{-6}=4 \times 10^{-8} \mathrm{~F}$ Now, given potential difference across this combination, $V=100 \mathrm{~V}$ So, charge on combination $ \begin{aligned} Q & =C_{\mathrm{eq}} V=4 \times 10^{-8} \times 100 \\ & =4 \times 10^{-6} \mathrm{C} \end{aligned} $

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

Practice more Electrostatics questions on Aicharya