Question

A capacitor of capacitance $C = 2.0 \pm 0.1\mu F$is charged to a voltage $V = 20 \pm 0.2V$. The charge $Q$ on the capacitor is then,

A. $40 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}$
B. $50 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}$
C. $60 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}$
D. $80 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}$

Answer 1

Hint:First, we should use the equation of charge in a capacitor, $Q = CV$ to calculate the charge $Q$ of the capacitor having voltage $V$.

Complete step by step solution:
The magnitude of capacitance and voltage is given in the question and the charge can be calculated by multiplying voltage and capacitance.
The error can be calculated separately using $\dfrac{{\Delta Q}}{Q} = \dfrac{{\Delta C}}{C} + \dfrac{{\Delta V}}{V}$.
Given,
The capacitance in the capacitor, $C = 2.0 \pm 0.1\mu F$.
The voltage of the capacitor, $V = 20 \pm 0.2\,V$.
The magnitude of capacitance, $C = 2.0\mu F$.
The error in capacitance, $\Delta C = 0.1\mu F$.
The error in voltage, $\Delta V = 0.2{\rm{ }}V$.
The error in charge is given by $\Delta Q$.
The error is given in terms of small change $\left( \Delta \right)$ in the magnitude..
The capacitance tells us how much charge the device stores for a given voltage. A dielectric between the conductors increases the capacitance of a capacitor.
The capacitance C is the proportional constant, and the equation of charge of capacitor is given by,
$Q = CV$ …… (1)

The given values of capacitance and voltage has small change or error of $ \pm $ .
The equation to find error of charge is given by
$\dfrac{{\Delta Q}}{Q} = \dfrac{{\Delta C}}{C} + \dfrac{{\Delta V}}{V}$ …… (2)

Substituting the value $2.0{\rm{ }}\mu F$in $C$and $2{\rm{ }}V$in $V$in equation (1) we get,
$\begin{array}{c}Q = 2 \times 20\\ = 40{\rm{ Coulomb}}\end{array}$

Substituting the value ${\rm{0}}{\rm{.1 \mu F}}$in $\Delta C$and $0.2{\rm{ V}}$in $\Delta V$in equation (2) we get,

$\begin{array}{l}\dfrac{{\Delta Q}}{{40}} = \dfrac{{0.1 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 6}}}} + \dfrac{{0.2}}{{20}}\\ \Rightarrow \Delta Q = 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}\end{array}$

The charge $Q$in the capacitor is,
$Q = 40 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}$

Hence, the correct answer is (A).

Note:In the solution, the students have to first know the relation of charge and voltage with the capacitance of a capacitor. The magnitude of the charge can be calculated by multiplying the value of capacitance and voltage. Since the values of voltage and charge have $ \pm $ error, hence this error can be considered as a small deviation or change in its value and the error in charge can be calculated from the error value given for capacitance and voltage.