Electric Fields - Comprehensive Notes

Electric Fields

Unit Overview

• Teaching Periods: 16

• Weightage: 11%

• Objectives:

  • Define Electrostatic force

  • Explain Coulomb's law

  • Define the Coulomb's force in different mediums

  • Solve problems using Coulomb's law

  • Describe the concept of an electric field as an example of a field of force.

  • Derive the expression $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$ for the magnitude of electric field at a distance 'r' from a point charge 'q'.

  • Define electric field strength as a force part unit positive charge.

  • Solve problems and analyze information using $\vec{E} = \frac{\vec{F}}{q_0}$.

  • Solve problems involving the use of the expression, $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$

  • Describe the concept of electric dipole

  • Calculate the magnitude and direction of the electric field at a point due to two charges with the same or opposite signs.

  • Sketch the electric field lines for two point-charges of equal magnitude with the same or opposite signs.

  • Describe electric flux.

  • Explain electric flux through a surface enclosing a charge.

  • Define absolute ele

    int charge using the equation, $V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$

  • Show that the electric field at a point is given by the negative potential gradient at that point.

  • Solve problems by using the expression, $E = -\frac{V}{d}$

  • Define electron volt.

8.1.1 Electrostatic Force

• Early Greek philosophers observed electrostatic phenomena with amber and dust particles.

• Electrostatic force is the attraction between dust particles and amber when amber is rubbed with silk or wool.

• Electrostatic force can be attractive or repulsive, unlike gravitational force, which is always attractive.

• Electrostatic forces are short-range compared to gravitational forces but stronger.

• Objects with the same charge repel, and objects with opposite charges attract.

8.1.2 Coulomb's Law

• Charles-Augustin de Coulomb used a torsion balance in 1785 to measure electric forces between charged objects.

• Gravitational force is negligible compared to the electric force between charged spheres.

• Electric Force Properties:

  • Charged particles exert an electrostatic force on each other.

  • The direction of force depends on the signs of the charges.

  • Like charges repel; force vectors point away from each other.

  • Unlike charges attract; force vectors point towards each other.

• The electrostatic force $F$ is:

  • Directly proportional to the product of the magnitudes of the charges $q<em>1$ and $q</em>2$.

  • Inversely proportional to the square of the distance $r$ between the charges.

  • Expressed as:

    • $F \propto \frac{q<em>1 q</em>2}{r^2}$

  • This relationship is known as the inverse square law.

• Coulomb's Law Equation:

  • $F = k \frac{q<em>1 q</em>2}{r^2}$

  • Where $k$ is Coulomb's constant.

  • $k = \frac{1}{4\pi\epsilon_0}$

  • The equation can be written as:

    • $F = \frac{1}{4\pi\epsilon<em>0} \frac{q</em>1 q_2}{r^2}$

8.1.3 Coulomb's Force in Different Mediums

• The permittivity of air is approximately 1.005 times that of free space ($\epsilon_0$), often considered equal.

• The permittivity of water is about 80 times that of a vacuum, reducing the force between charges in water.

• General Equation:

  • $F = \frac{1}{4\pi\epsilon r^2} q<em>1 q</em>2$

  • Where $\epsilon = \epsilon<em>r \epsilon</em>0$ is the permittivity of the medium.

  • $\epsilon_r$ is the relative permittivity.

• Permittivity of Free Space:

  • $\epsilon_0 = 8.854 \times 10^{-12} \frac{C^2}{N \cdot m^2}$

• Coulomb's Constant in SI Units:

  • $k = 8.988 \times 10^9 \frac{N \cdot m^2}{C^2}$

• Salt crystals dissolve in water because the electrostatic forces between sodium and chlorine ions are overcome due to water's high permittivity.

• Coulomb's law strictly applies to a vacuum; in air, the force is effectively the same.

Vector Form of Coulomb's Law

• The electric force $\vec{F}<em>{12}$ exerted by charge $q</em>1$ on $q_2$ is:

  • $\vec{F}<em>{12} = \frac{1}{4\pi\epsilon</em>0} \frac{q<em>1 q</em>2}{r^2} \hat{r}_{12}$

  • $\hat{r}<em>{12}$ is a unit vector directed from $q</em>1$ to $q_2$.

• Electric force obeys Newton's third law:

  • $\vec{F}<em>{12} = -\vec{F}</em>{21}$

Worked Example 8.1

• Problem: Find the electric and gravitational forces between an electron and proton in a hydrogen atom, separated by $5.3 \times 10^{-11} m$.

• (a) Electric Force Calculation:

  • $F<em>e = k \frac{q</em>e q_p}{r^2} = 8.988 \times 10^9 \frac{N \cdot m^2}{C^2} \times \frac{(-1.602 \times 10^{-19} C)(1.602 \times 10^{-19} C)}{(5.3 \times 10^{-11} m)^2}$

  • $F_e = -8.2 \times 10^{-8} N$ (attractive)

• (b) Gravitational Force Calculation:

  • $F<em>g = G \frac{m</em>e m_p}{r^2} = 6.67 \times 10^{-11} \frac{N \cdot m^2}{kg^2} \times \frac{(9.109 \times 10^{-31} kg)(1.672 \times 10^{-27} kg)}{(5.3 \times 10^{-11} m)^2}$

  • $F_g = 3.6 \times 10^{-47} N$

• (c) Conclusion: The gravitational force is negligible compared to the electric force.

Worked Example 8.2

• Problem: Two positive charges repel each other with a force of 0.1 N when 50 cm apart in a vacuum. Find the charge value and the force if placed in an insulating liquid with $\epsilon = 5\epsilon_0$.

• (a) Charge Calculation:

  • $F = k \frac{q^2}{r^2} \Rightarrow q^2 = \frac{F \times r^2}{k} = \frac{0.1 N \times (0.5 m)^2}{8.988 \times 10^9 \frac{N \cdot m^2}{C^2}}$

  • $q = 1.7 \times 10^{-7} C = 0.17 \mu C$

• (b) Force in Liquid Calculation:

  • $F_{liquid} = \frac{F}{5} = \frac{0.1 N}{5} = 0.02 N$

Worked Example 8.3

• Problem: Three charges $q<em>1 = +2 \mu C$, $q</em>2 = +3 \mu C$, and $q<em>3 = +4 \mu C$ are at the vertices of an equilateral triangle with sides of 10 cm. Calculate the resultant force on $q</em>3$.

• Solution:

  • $F<em>1 = k \frac{q</em>1 q_3}{r^2} = 8.988 \times 10^9 \frac{N \cdot m^2}{C^2} \times \frac{(2 \times 10^{-6} C)(4 \times 10^{-6} C)}{(0.1 m)^2} = 7.2 N$

  • $F<em>2 = k \frac{q</em>2 q_3}{r^2} = 8.988 \times 10^9 \frac{N \cdot m^2}{C^2} \times \frac{(3 \times 10^{-6} C)(4 \times 10^{-6} C)}{(0.1 m)^2} = 10.8 N$

  • Using components and resolving forces:

    • $F = \sqrt{F<em>1^2 + F</em>2^2 + 2 F<em>1 F</em>2 (\sin^2 \theta - \cos^2 \theta)}$

    • $F = \sqrt{(7.2)^2 + (10.8)^2 + 2(7.2)(10.8)((sin 60^\circ)^2 - (cos 60^\circ)^2)} = 15.69 N$

Self-Assessment Questions

1. Calculate the separation between two electrons in a vacuum for which the electric force equals the gravitational force on one of them at the Earth's surface.

2. Show that the force between charges decreases when a dielectric medium is filled between them compared to when they are in air.

8.2.1 Electric Field

• Michael Faraday introduced the concept of the electric field.

• An electric field exists in the space around a charged object (source charge).

• Electric field lines are imaginary but can be visualized by the motion of a test charge.

• Field lines indicate the field strength and direction.

• Field lines are radial, do not intersect, originate from positive charges, and terminate on negative charges.

• Electric field is strong where field lines are close together.

• Sensitive electronic devices are often enclosed in metal boxes to eliminate stray electric field interference.

Strength of Electric Field

• A test charge experiences an electric force due to the source charge's field.

• Electric Field $\vec{E}$ is defined as the electric force $\vec{F}$ on a positive test charge $q_0$ divided by the test charge:

  • $\vec{E} = \frac{\vec{F}}{q_0}$

• SI unit of electric field: volts per meter (V/m) or newtons per coulomb (N/C).

8.2.2 Electric Field of Point Charge

• A point charge $q$ creates an electric field around it.

• A test charge $q_0$ is placed at a distance $r$ from the source charge.

• The test charge is small enough not to affect the field of the source charge.

• The force exerted on the test charge is given by Coulomb's law:

  • $\vec{F} = \frac{1}{4\pi\epsilon<em>0} \frac{q q</em>0}{r^2} \hat{r}$

• Electric field strength is defined as the electric force on the test charge per unit charge:

  • $\vec{E} = \frac{\vec{F}}{q<em>0} = \frac{1}{4\pi\epsilon</em>0} \frac{q}{r^2} \hat{r}$

• Magnitude of the electric field at a distance $r$:

  • $E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$

  • $E \propto \frac{1}{r^2}$

• Electric field intensity decreases with distance from the source charge.

• If there are $n$ charged particles, the net electric field is the vector sum of individual electric fields:

  • $\vec{E} = \vec{E}<em>1 + \vec{E}</em>2 + \vec{E}<em>3 + … + \vec{E}</em>n$

Worked Example 8.4

• Problem: Find the electric field intensity where a proton experiences a force equal to its weight.

• Solution:

  • $E = \frac{F}{q<em>p} = \frac{mg}{q</em>p} = \frac{(1.672 \times 10^{-27} kg)(9.8 m/s^2)}{1.602 \times 10^{-19} C}$

  • $E = 1.022 \times 10^{-7} N/C$

Worked Example 8.5

• Problem: Calculate the magnitude of the electric field if a test charge of $3.5 \mu C$ experiences a force of 70 mN. If the test charge is replaced by an electron, calculate the force on the electron.

• (a) Electric Field Calculation:

  • $E = \frac{F}{q_0} = \frac{70 \times 10^{-3} N}{3.5 \times 10^{-6} C} = 2.0 \times 10^4 N/C$

• (b) Force on Electron Calculation:

  • $F = Eq_e = (2.0 \times 10^4 N/C)(1.602 \times 10^{-19} C) = 3.204 \times 10^{-15} N$ (upwards)

Self-Ale

• An electric dipole consists of two opposite electric charges of equal magnitude, separated by a small distance $d$.

• The dipole moment $\vec{p}$ is a vector representing the strength and direction of the electric dipole.

  • $p = q \times d$

8.3.2 Electric Field at Point Due to Two Charges

• Consider charges “-q” and “+q” separated by a distance “d”.

• Resolve field vectors into components.

• The net electric field is the vector sum of the horizontal components.

• Magnitude of both electric fields is the same:

  • $E<em>+ = E</em>- = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$

  • $E = 2 \times \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \cos\theta$

• From triangle ADC, $\cos\theta = \frac{d/2}{r}$, where $r = \sqrt{y^2 + (d/2)^2}$.

  • $E = \frac{1}{4\pi\epsilon_0} \frac{qd}{(y^2 + (d/2)^2)^{3/2}}$

• The electric dipole moment is the product of the charge “q” and “d”.

  • $\vec{p} = q \vec{d}$

• Dipole moment is directed from the negative to the positive charge.

• If d << y,

  • $E = \frac{1}{4\pi\epsilon_0} \frac{p}{y^3}$

Worked Example 8.6

• Calculate the electric field intensity at point P due to two positive charges of the same magnitude separated by a small distance “d”.

• The net electric field is due to the vector sum of the vertical components.

  • $E_{total} = 2E \sin\theta$

  • Since $\sin\theta = \frac{y}{r}$, and $r = \sqrt{y^2 + (d/2)^2}$

  • $E<em>{total} = \frac{1}{2\pi\epsilon</em>0} \frac{qy}{(y^2 + (d/2)^2)^{3/2}}$

• If d < y,

  • $E = \frac{1}{2\pi\epsilon_0} \frac{q}{y^2}$

8.4.1 Electric Flux

• Consider an electric field that is uniform in both magnitude and direction.

• Total number of lines penetrating the surface is proportional to the dot product of $\vec{E} \cdot \vec{A}$.

• Electric flux is denoted by $\Phi_e$:

  • $\Phi_e = \vec{E} \cdot \vec{A}$

• SI unit of electric flux is $N \cdot m^2/C$.

Cases

• If the vector area $\vec{A}$ is parallel to the field, $\Phi_e = EA$

• If the vector area $\vec{A}$ is perpendicular to the field, $\Phi_e = 0$

• If the vector area $\vec{A}$ is antiparallel to the field, $\Phi_e = -EA$

8.4.2 Electric Flux Through a Surface Enclosing a Charge

• Divide the surface into $N$ small elements of area $\Delta A_i$.

• Net electric flux through the closed surface is given as:

  • $\Phi<em>e = \sum</em>{i=1}^{N} \vec{E} \cdot \Delta \vec{A}_i$

• The electric flux through a sphere can be calculated by considering a sphere of radius