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Question: A capacitor of capacitance \(C = 2.0 \pm 0.1\mu F\)is charged to a voltage \(V = 20 \pm 0.2V\). The ...

A capacitor of capacitance C=2.0±0.1μFC = 2.0 \pm 0.1\mu Fis charged to a voltage V=20±0.2VV = 20 \pm 0.2V. The charge QQ on the capacitor is then,

A. 40±2.4×10−6Coulomb40 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}
B. 50±2.4×10−6Coulomb50 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}
C. 60±2.4×10−6Coulomb60 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}
D. 80±2.4×10−6Coulomb80 \pm 2.4 \times {10^{ - 6}}{\rm{ Coulomb}}

Explanation

Solution

First, we should use the equation of charge in a capacitor, Q=CVQ = CV to calculate the charge QQ of the capacitor having voltage VV.

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 ΔQQ=ΔCC+ΔVV\dfrac{{\Delta Q}}{Q} = \dfrac{{\Delta C}}{C} + \dfrac{{\Delta V}}{V}.
Given,
The capacitance in the capacitor, C=2.0±0.1μFC = 2.0 \pm 0.1\mu F.
The voltage of the capacitor, V=20±0.2 VV = 20 \pm 0.2\,V.
The magnitude of capacitance, C=2.0μFC = 2.0\mu F.
The error in capacitance, ΔC=0.1μF\Delta C = 0.1\mu F.
The error in voltage, ΔV=0.2V\Delta V = 0.2{\rm{ }}V.
The error in charge is given by ΔQ\Delta Q.
The error is given in terms of small change (Δ)\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=CVQ = 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
ΔQQ=ΔCC+ΔVV\dfrac{{\Delta Q}}{Q} = \dfrac{{\Delta C}}{C} + \dfrac{{\Delta V}}{V} …… (2)

Substituting the value 2.0μF2.0{\rm{ }}\mu Fin CCand 2V2{\rm{ }}Vin VVin equation (1) we get,
Q=2×20 =40Coulomb\begin{array}{c}Q = 2 \times 20\\\ = 40{\rm{ Coulomb}}\end{array}

Substituting the value 0.1μF{\rm{0}}{\rm{.1 \mu F}}in ΔC\Delta Cand 0.2V0.2{\rm{ V}}in ΔV\Delta Vin equation (2) we get,

ΔQ40=0.1×10−62×10−6+0.220 ⇒ΔQ=2.4×10−6Coulomb\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 QQin the capacitor is,
Q=40±2.4×10−6CoulombQ = 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.