Exercise EE1-01.03 – First electrical quantities and formulae

Prof. Dr. Thorsten Jungmann (last updated 2025-09-12)

Relates to EE1-01 Introduction

Questions/Tasks

a) Which electrical engineering quantities were introduced in this lesson?
b) Which mathematical formulae were introduced? What do they state?
c) Convert the following amounts of electric charge into the units given:

• $2.5\cdot 10^{12}$ elementary charges = ?? $\mathrm{C}$
• $1.6\mathrm{mC}$ = ?? elementary charges
• $-850\mathrm{nC}$ = ?? $\mathrm{C}$
• $0.0024\mathrm{C}$ = ?? $\mathrm{mC}$ = ?? $\mu \mathrm{C}$

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Solution

a) Quantities

Electric charge $Q$; electric charge of the electron: $-e\approx -1.602\cdot 10^{-19}\mathrm{C}$
Force on electric charges $F$
Coulomb constant $k=\frac{1}{4\pi {\epsilon }_0}\approx 8.988\cdot 10^9\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{C}}^2$
Electric field constant ${\epsilon }_0=8.854\cdot 10^{-12}F/m$
Electric field strength $E$ in $\frac{\mathrm{V}}{\mathrm{m}}$

b) Formulae and what they state

Maxwell’s equations

• Electric charges generate electric fields. These exert forces on mobile charge carriers located in them and set them in motion.
• Moving charge carriers generate magnetic fields.
• Changing electric fields generate magnetic fields.
• Magnetic fields influence the motion of charge carriers.

Coulomb force

$$F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{|Q_1\cdot Q_2|}{r^2}$$

Direction of the force:

• Like charges (both + or both −): the force is repulsive
• Opposite charges (+ and −): the force is attractive

Proportionalities:

• $F\sim Q_1\cdot Q_2$ (proportional to the product of the electric charges)
  • The more electric charge, the stronger the force
• $F\sim \frac{1}{r^2}$ (inversely proportional to the square of the distance)
  • The greater the distance, the weaker the force

Electric field strength

The electric field strength $E$ in $\frac{\mathrm{V}}{\mathrm{m}}$ describes how strong the electric field is at a given point and thus characterizes the spatial distribution of the force that an electric field exerts on an electric charge. The greater the electric field strength, the greater the force on the electric charge $q$.

$$F=q\cdot E$$

The electric field strength between the plates (plate separation $d$) of a capacitor that has been charged to the voltage $U$:

$$E=\frac{U}{d}$$

c) Conversions

General formulae:
Conversion between elementary charges and coulombs:

$$Q_{\mathrm{C}}=n\cdot e\text{or}n=\frac{Q_{\mathrm{C}}}{e}$$

where:

• $Q_{\mathrm{C}}$ is the electric charge in coulombs ($\mathrm{C}$)
• $n$ is the number of elementary charges
• $e=1.602\cdot 10^{-19}\mathrm{C}$ is the elementary charge

Unit conversion with powers of ten:

$$1\mathrm{C}=1\cdot 10^3\mathrm{mC}=1\cdot 10^6\mu \mathrm{C}=1\cdot 10^9\mathrm{nC}$$

Solutions:

a) $Q=n\cdot e=2.5\cdot 10^{12}\cdot 1.602\cdot 10^{-19}\mathrm{C}=4.0\cdot 10^{-7}\mathrm{C}=\underline{0.40\mu \mathrm{C}}$

b) $n=\frac{Q}{e}=\frac{1.6\cdot 10^{-3}\mathrm{C}}{1.602\cdot 10^{-19}\mathrm{C}}=9.99\cdot 10^{15}\approx \underline{1.0\cdot 10^{16}}$ elementary charges

c) $-850\mathrm{nC}=\underline{-850\cdot 10^{-9}\mathrm{C}}$

d) $0.0024\mathrm{C}=2.4\cdot 10^{-3}\mathrm{C}=\underline{2.4\mathrm{mC}}=\underline{2400\mu \mathrm{C}}$