Question

What charge is stored in a $\mathbf{180}\mathbf{\mu F}$capacitor when 120Vis applied to it?

Short answer

The amount of charge stored in a $180\mathrm{\mu F}$capacitor when 120V is applied to it, is $\mathrm{Q}=21.6\mathrm{mC}$.

Step-by-step solution

Given Data

Capacitance of capacitor is$\mathrm{C}=180\mathrm{\mu F}$

Potential difference is V = 120V

Concept Introduction

Capacitors: A pair of conductors separated by a dielectric, when charged, charges of equal magnitude Q and opposite sign accumulate on the two conductors, and the potential so developed $\mathbf{∆}\mathbf{V}$is directly proportional to the capacitance if charge is kept constant. The capacitance is given as-

$\mathbf{C}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{\Delta V}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Calculation for Charge Stored

The charge stored in the capacitor is found by solving Equation (1) for Q–

$\mathbf{Q}\mathbf{=}\mathbf{C\Delta V}$

Entering the values for C and$∆\mathrm{V}$, it is obtained –

$\begin{array}{rcl}\mathbf{Q}& \mathbf{=}& \left(180\times {10}^{-6}F\right)\mathbf{(}\mathbf{120}\mathbf{}\mathbf{V}\mathbf{)}\\ & \mathbf{=}& \mathbf{21}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{C}\\ & \mathbf{=}& \left(21.6\times {10}^{-3}C\right)\left(\frac{{10}^3\mathrm{mC}}{1C}\right)\\ & \mathbf{=}& \mathbf{21}\mathbf{.}\mathbf{6}\mathbf{mC}\end{array}$

Therefore, the charge stored is $\mathbf{Q}\mathbf{=}\mathbf{21}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{mC}$.