TOPIC 6.2: Electric Fields and Forces
6.2.1 State that there are two types of electric charge.
 Have you ever removed a wool sweater and heard a strange crackling noise?
 Have you ever opened a band aid and seen a flash?
 Whenever two objects rub against each other there is friction. This friction can cause “electrification” effect
as tiny charged particles are forced to move form one region to another.
 These charged particles are electrons.
 Negative charge
 Outside of atom
 Can easily be stripped away by friction.
 “Orbit” the nucleus.
 The nucleus contains
 Protons, positive charge
 Neutrons, no charge.
 UNITS
 Electric charge = q or Q. Measured in Coulombs/C
 1 Coulomb of charge = the charge carried on 6.25 x 1018 electrons
OR
 The charge on 1 electron(e) = 1.6 x 10-19Coulombs
IB GIVEN DATA
 Charge on electron – e = 1.60 x 10-19C
 Electron rest mass – me = 9.11x10-31kg = 0.000549u = 0.511MeV/c2
 Proton rest mass – mp = 1.673x10-27kg = 1.007276u = 938MeV/c2
 Neutron rest mass – mm = 1.675x10-27kg = 1.008665u = 940MeV/c2
 Two types of charges:
 Positive and negative.
 Uncharged/neutral objects have equal amounts of positive and negative charges.
 Proton and electron have equal amounts of charge
 But proton is about 1800 times more massive.
 Electrification occurs because electrons move.
 This creates a negative area and leaves behind an equally positive region.
 Net charge is produced when there is an imbalance in the number of protons and electrons.
 Some objects naturally become positively charged. Others naturally become negatively charged.
 Example: Rub an acrylic plastic ruler on a piece of cloth. The friction between the plastic and the cloth
causes some of the electrons to be stripped from the surface of the plastic and deposite don the wool. This
gives an overall positive charge to the plastic and a negative charge to the cloth.
6.2.2 State and apply the law of conservation of charge.
IB Definition
 Charge cannot be created or destroyed but it can move from one place to another or from one object
to another.
 Charges can move from one place to another.
 Charges can NOT be created from nothing.
Practice 6
 A metal sphere carrying a charge of +4.2µC comes into contact with an identical second sphere which is
neutral. Calculate the charge on each sphere.
 Answer = 2.1µC
Practice 7
 Two identical metal spheres carry charges of –3.2µC and –4.8µC respectively. The two spheres are brought
together and the charge evenly distributes itself so that each sphere carries equal charge. Determine the
charge on each sphere.
 Answer = –4µC
Practice 8
 Two metal spheres of different size carry charges of –3.0µC and +4.8µC respectively. The two spheres are
brought together and the charge distributes itself so that one sphere carries twice as much charge as the
second sphere. Determine the charge on each sphere.
 Answer = +1.2µC and +0.6µC
6.2.3 Describe and explain the difference in the electrical properties of conductors and insulators.
 Conductors – materials that allow a free movement of electrons both within the substance and on its outer
surface.
 Insulators – do not allow electrons to move freely. Do allow the build-up of electrons over its outer surface.
 If a charged insulator is held, it will keep its charge.
 If a conductor is held it will lose its charge. Electrons will flow from the conductor to the person and then to
the earth.
 Rubber soled shoes don’t allow this.
 Examples of Conductors
 All metals – gold, silver, copper, ect.
 Carbon in the form of graphite
 Examples of Isulators
 Plastics – perspex, poythene, PVC
 Rubber
 Wood
 Glass
6.2.4 State Coulomb’s Law.
 Opposites attract/likes repel ***See diagram on board***
 Two charged objects are brought together and an electrical force is created between them.
 Size of force depends upon size of each charge and the distance between them.
 This electrical force is directly proportional to the product of the two charges.
 This electrical force is inversely proportional to “the square of the distance between them”.
IB Definitions/Data
 Coulomb’s Law – the force between two point charges is directly proportional to the product of their
charges and inversely proportional to the square of the distance between them. The direction of the
force is along the line connecting the two charges.
 Coulomb constant = k = 8.99 x 109 Nm2/C2
 Permittivity of free space = ε0 = 8.85 x 10-12 C2/Nm2
 F = kq1q2/r2 = 1 / 4πε0 * q1q2/r2 (write this out)
 F ∝ q1q2/r2
 F = kq1q2/r2
 k = a constant of proportionality called Coulomb’s constant
 k = 1/4πε0
 ε0 = permittivity of free space
 ε0 = 1/4πk
Practice 9
 A +4µC charge is placed 20cm away from a –6µC charge. Calculate the force between these charges. Is this
an attractive or repulsive force?
 Answer: - 5.39 Newtons
 A +4µC
6.2.5 Define electric field strength.
 We’ve already looked at the idea of a filed when we looked at gravitational fields.
 A field is a region of space where a force acts without contact.
 Around any electric charge, there exists an “electric field”
 A region of space where a force would exist if another charge entered the field.
IB Definition
 Electric Field – the area around an electric charge where a force exists if another charge enters the
field.
 Point Charge
 We use the idea of a “test charge” or “point charge” to investigate electric field strength.
 A test charge is considered to be a positive charge whose charge is so small that I does not alter the
electric field in any way. (The charge can be ignored.)
 Check out this simulation to better under stand a test charge.
 http://phet.colorado.edu/en/simulation/electric-hockey
 Field strength of the electric field is proportional to the magnitude of the force at that point and inversely
proportional to the charge. Also notice the units are N/C or NC-1
IB Definition: E = F/q
 Electric Field Strength(E) – the force per unit charge on a tiny positive test charge.
 E = F/q → F = Eq
 And since F = kq1q2/r2
 Eq = kq1q2/r2
 E= k(q/r2)
6.2.6 Determine the electric field strength due to one or more point charges.
Problem 10
 Calculate the strength of the electric field 0.4M away form a point charge of magnitude +20µC.
 Answer: E=1.12 x 106 N/C
 Remember that electric field has both magnitude and direction. This means that we can use vector addition
to solve of the net electric field.
Problem 11
 The arrangement shows three point charges q1, q2, and q3. The sizes of these charges are +6µC, +4µC and 3µC respectively. They are separated by the distances shown.
 Calculate the magnitude and direction of the resulting force on q1
 Answer 21.1N @ 71.6º
Problem 12
 Calculate the magnitude and direction of the electric field for a test charge placed at the position shown due
to the two point charges q1 and q2 of magnitude +5µC and -8µC. The distance from the test charge to the
two individual charges is shown.
 Answer: 142 x 103 N/C @ 62º
6.2.7 Draw the electric field patterns for different charge configurations.
 Radial field - The electric field lines are directly towards or away from the point charge.
 The closer the field line the larger the relative magnitude of the field.
 As the distance away fro the charge increases, the strength of the field diminishes and the gap between the
field lines increases.
 Notice electric field lines are directed away
from the positive charge and toward the
negative.
 Which charge is the strongest? Weakest?
 Equipotential Lines
 An equipotential is a line or surface over which the potential is constant.
 Electric field lines are perpendicular to equipotentials.
 The surface of a conductor is an equipotential.
 Guidelines for drawing field lines:
 Electric field lines never cross each other.
 Electric field lines are most dense around objects with the greatest amount of charge.
 At locations where electric field lines meet the surface of an object, the lines are perpendicular to the
surface.
6.2.8 Solve problems involving electric charges, forces and fields..
 ***See previous problems***
Problem 16
 Erin Agin drew the following electric field lines for a configuration of two charges. What did Erin do
wrong? Explain.
Problem 17
 Consider the electric field lines shown in the diagram below.
From the diagram, it is apparent that object A is ____ and
object B is ____.
Problem 18
 Consider the electric field lines drawn at the right for a configuration of two charges. Several locations are
labeled on the diagram. Rank these locations in order of the electric field strength - from smallest to largest.
 Answer: DAECB (with the order of C and B being in question)