Question

How many$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\mu F}$capacitors must be connected in parallel to store a charge of 1.00 Cwith a potential of 110 Vacross the capacitors?

Short answer

he number of capacitors that need to be connected in parallel is$\mathrm{N}=9.1\times {10}^3$

Step-by-step solution

Given data

The capacitance is$\mathrm{C}=1.00\mathrm{\mu F}$

Charge to store is q = 1.00 C.

The potential across the capacitors is V = 110 V.

Determining the concept

Combining the formula for the equivalent capacitance of Nidentical capacitors in parallel, and (equation 25-1), find the number of capacitors required to store a capacity 1.00 C.

Formulae are as follows:

$$\begin{gathered} \mathbf{C}\mathbf{=}\frac{\mathbf{q}}{\mathbf{V}} \\ {\mathbf{C}}_{\mathbf{eq}}\mathbf{=}\mathbf{NC} \end{gathered}$$

Where C is capacitance, V is the potential difference, q is the charge on particle, and N is no. Of capacitors.

Determining the number of capacitors that need to be connected in parallel

The equivalent capacitance is given by,

${\mathrm{C}}_{\mathrm{eq}}=\frac{\mathrm{q}}{\mathrm{V}}$

Where, q is the total charge on all capacitors, and V is the potential difference across any one of them. For N identical capacitors in parallel,

${\mathrm{C}}_{\mathrm{eq}}=\mathrm{NC}$

Where, C is the capacitance of one of the capacitors.

Thus,

$\mathrm{NC}=\frac{\mathrm{q}}{\mathrm{V}}$

Therefore, the number of capacitors is,

$$\begin{gathered} \mathrm{N}=\frac{\mathrm{q}}{\mathrm{VC}} \\ \mathrm{N}=\frac{1.00\mathrm{C}}{(110\mathrm{V})\times (1.00\times {10}^{-6}\mathrm{F})} \\ =9.1\times {10}^{-3} \end{gathered}$$

Hence, the number of capacitors that need to be connected in parallel is$\mathrm{N}=9.1\times {10}^3$.

Therefore, by combining the formula for the equivalent capacitance of N identical capacitors in parallel the required capacitors can be determined.