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 q1 and q2.

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

  • Expressed as:

    • F \propto \frac{q1 q2}{r^2}

  • This relationship is known as the inverse square law.

• Coulomb's Law Equation:

  • F = k \frac{q1 q2}{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\epsilon0} \frac{q1 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} q1 q2

  • Where \epsilon = \epsilonr \epsilon0 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}{12} exerted by charge q1 on q_2 is:

  • \vec{F}{12} = \frac{1}{4\pi\epsilon0} \frac{q1 q2}{r^2} \hat{r}_{12}

  • \hat{r}{12} is a unit vector directed from q1 to q_2.

• Electric force obeys Newton's third law:

  • \vec{F}{12} = -\vec{F}{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:

  • Fe = k \frac{qe 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:

  • Fg = G \frac{me 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 q1 = +2 \mu C, q2 = +3 \mu C, and q3 = +4 \mu C are at the vertices of an equilateral triangle with sides of 10 cm. Calculate the resultant force on q3.

• Solution:

  • F1 = k \frac{q1 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

  • F2 = k \frac{q2 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{F1^2 + F2^2 + 2 F1 F2 (\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\epsilon0} \frac{q q0}{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}}{q0} = \frac{1}{4\pi\epsilon0} \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}1 + \vec{E}2 + \vec{E}3 + … + \vec{E}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}{qp} = \frac{mg}{qp} = \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+ = E- = \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{total} = \frac{1}{2\pi\epsilon0} \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:

  • \Phie = \sum{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