Strand G. Electricity Unit 1. Electrostatics Contents Page Charge 2 Forces Between Charges 4 Electric Field 10
G.1.1. Charge An atom was once considered the building block of matter, although we know now that it is not the smallest most basic building block. As we will learn in Strand J, all atoms consist of a central nucleus, containing positively charged protons (p) and uncharged neutrons (n). Orbiting in the space surrounding the nucleus are electrons (e). These sub atomic particles have an electrical charge equal in magnitude to the proton, but of opposite sign. Thus in the following unit, charge refers to either electrons or protons (but mostly electrons since it is the electron that is the charged particle responsible for electrical flow through a wire). The property of electrical charge is as fundamental as mass, and just like objects with mass are accelerated by the force of gravity, objects with charge are accelerated by electric force. The amount of charge is measured in Coulombs (C), and the charge of the sub atomic particles constituting the atom are shown in Table G.1.1.1, together with their respective mass. Table G.1.1.1. Particle Charge (C) Mass (kg) Proton (p) 1.6 10 19 1.67 10 27 Neutron (n) 0 1.67 10 27 Electron (e) 1.6 10 19 9.1 10 31 Since the proton and the electron have such a tiny charge as measured in Coulombs, and the charge of an electron or proton is a natural unit of charge (any charged system has a total charge that is a multiple of 1.6 10 19 ) the relative charge is often assigned such that the charge of a proton is 1, a neutron 0, and an electron 1. In this way a charge of 2 is equal to 2 (1.6 10 19 ) = 3.2 10 19 C. Single charges are usually denoted by either the simple q, which can be either negative or positive, or by the particle symbol (p for proton and e for the electron). Overall charge, which tends to be a large number of individual charges, is denoted by the symbol Q. The atoms in a material or element are usually neutral, having the same number of protons and neutrons in each atom. However, atoms are able to swap or share electrons, although they cannot share protons or neutrons. If an atom loses an electron it becomes positively charged, and we call this and ion. If an electron is added to an atom it becomes negatively charged. This ability of atoms to share or swap electrons provides an opportunity to classify materials and elements into three groups, conductors, semiconductors, and insulators. The atoms in a conductor, such as a metal, have electrons that are loosely bound to the nucleus. These spare loosely bound atoms are free to hop from one atom to another. These materials are conducting, since the electrons are free to move through the material, conveying charge (conduct electricity). The electrons of a semiconductor are more tightly bound to their parent atoms, and at lower temperatures, are not conducting. However, when heated the 2
electrons have enough energy to break free from the atoms and move through the material, and the semiconductor becomes conducting. Insulators, such as plastics, have electrons that are very tightly bound to atoms. In addition, the material would melt before any electrons could be freed by increasing the temperature of the material. Insulators do not conduct electricity since they have no free charges to do so. The interaction between charges that are at rest is known as electrostatics. When you rub a balloon and stick the balloon to your jumper or to a wall, this is a demonstration of electrostatics. Like the balloon and a jumper, plastic, fur and cloth are good materials for demonstrating electrostatic because they exchange electrons easily. Rubbing a polythene rod with a dry cloth transfers electrons to the surface atoms of the rod. Since the polythene rod now has more electrons, it is more negative, or negatively charged. The dry cloth has given up electrons. It is therefore less negatively charged or positively charged relative to the rod. Note that being positively charged can mean an absence of negative charge. Polythene Cloth Perspex Cloth Figure G.1.1.1. Rubbing a Perspex rod with a dry cloth result in transfer of electrons from the rod to the surface atoms of the dry cloth. The Perspex rod gives up electrons and is therefore positively charged whilst the cloth gains electrons becoming negatively charged. The difference between these two scenarios is the material of the rod. Some insulators become negatively charged and some positively charged when rubbed, depending on how much the atoms of the material wants to gain or loose electrons. Perspex As a result of all the rubbing, we now have two Perspex insulating rods and two dry cloths that are charged. The Perspex rod is positively charged and the polythene rod negatively charged. If we Figure G.1.1.2. hang the Perspex rod up using an insulating thread such as cotton, as shown in Figure G.1.1.2, and bring the polythene rod towards it, we find that the two rods are attracted to each other. If we bring the negatively charged cloth up to the suspended Perspex rod we find that the rod and cloth repel each other. This allows us to conclude that; Two positive charges or two negative charges repel each other A positive charge and negative charge attract each other 3
This can easily be recalled by remembering like charges repel and opposites attract, although strictly speaking the term like charges does not necessarily refer to two identical particles with identical charge, but rather the algebraic sign of the charge for each is the same. Exercise G.1.1 1. Calculate the combined charge in Coulombs of four electrons. 2. In a system containing 9 electrons, 14 protons and 6 neutrons, what would be the total relative charge of the system? What would be the total charge in Coulombs? 3. Complete the following table for the proton (p) and electron (e) combinations. Particle Combination pp pe ee ep Total Relative Charge Total Charge (C) Force Between Particles 4. A polythene rod is charged by rubbing with a fur mitten. The charge on the rod is 2µC. How many electrons (to the nearest electron) were exchanged between the rod and the mitten? Challenge Question 5. Bob is at a party. For his party trick, (which his girlfriend Lauren finds very annoying) he rubs a balloon on his wooly jumper and holds it above Lauren s head, causing her hair to stand on end. Explain Bob s trick in terns of electrostatics. G.1.2. Forces Between Charges As like charges repel and opposites attract (recall that in this statement we are talking about the algebraic sign of the charge rather than type), there must be a force between the charges. Further, if the charges q1 and q2 are both positive or both negative, this force is repulsive. If one charge is positive whilst the other is negative, the force is attractive. Charles Augustin de Coulomb (17361806) studied the interaction forces between charged particles in detail in 1784, the unit of charge being named after him in recognition of his work. Coulomb found that for point charges (charges that are very small in comparison to the distance between them), the force between them was attractive or repulsive depending on the sign of the charge. In addition, he found that the magnitude of the force between charges reduced as the distance between the charges increased. Specifically, he found that the magnitude of the force reduced by a factor of four when ever he doubled the distance between the charges, an increased by a factor 4
of four whenever he halved the distance between the charges. The conclusion that Coulomb came to is now known as Coulombs Law. The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically Coulombs Law may be written; where FF = qq 1qq 2 4ππεε 0 rr 2 F = electrostatic force q = point charge (in Coulombs) r = distance between charges ε0 = a constant = 8.854 10 12 C 2 /Nm 2 Here the distance between the two charges r is always taken to be the shortest distance joining the center of the two charges, with the force acting along a straight line joining the two charges as shown in Figure G.1.2.1. The sign of F indicates the repulsive or attractive nature of the interaction, as shown by the following worked example. Worked Example F2 on 1 q1 q1 F2 on 1 r r q2 F1 on 2 F1 on 2 q2 Calculate the force that q1 exerts on q2 if Figure G.1.2.1. (a) q1 = 2µC and q2 = 3µC, and the two charges are separated by 10cm (b) q1 = 3nC and q2 = 4nC, and the two charges are separated by 0.4m Draw a free body diagram representing the direction of the force in each case. 5
Answer (a). Here we have two charges, q1 = 2µC = 2 10 6 C q2 = 3µC = 3 10 6 C 2µC 0.1m 3µC Separated by a distance of 0.1m as shown. The force that q1 exerts on q2 is found using Coulomb s Law; FF = qq 1qq 2 4ππεε 0 rr 2 = ( 2 10 6 CC)(3 10 6 CC) 4 3.14 8.85 10 12 0.1 2 = 5.4NN Since one charge is negative and the other positive, the product of the two charges is negative (a positive multiplied by a negative is always a negative), making F negative. This signifies that F is attractive, as expected from the opposites attract rule. Thus the force on q2 acts to the left. q1 5.4N 0.1m q2 From Newton s Third Law, the force acting on q1 is also 5.4N, but acts on the opposite direction (from left to right). (b). This time we have; q1 = 3nC = 3 10 9 C q2 = 4nC = 4 10 9 C 3nC 0.4m 4nC Separated by a distance of 0.4m as shown. The force that q1 exerts on q2 is found using Coulomb s Law; FF = qq 1qq 2 4ππεε 0 rr 2 = (3 10 9 CC)(4 10 9 CC) 4 3.14 8.85 10 12 0.4 2 = 6.75 10 7 NN Since both charges are positive, F is positive, and the force between the charges is repulsive. Therefore the force on q2 acts to the right. q1 0.4m q2 6.75 10 7 N Force is a vector quantity and as we learnt in Strand C, the superposition of forces must apply. For two charges that exert a force on a third charge, the resultant force acting on the third charge is the vector sum of the forces from the 6
other two charges. The superposition of forces applies to any number of charges. To solve a multiple charge problem; Draw a free body diagram of the charges and draw the direction of each force, being careful to evaluate if each individual force is attractive or repulsive. Calculate the magnitude of the force on the charge in question that is exerted from each of the other charges. Sum the forces acting vectorially, just as you would for vector mechanics, either by splitting the individual forces into their components (the easiest method) or by using the head to tail method. Worked Example A charge of 3µC is situated at the origin. A 2µC and 2µC charge are situated at coordinates (0.3,0) and (0.2,0.2) respectively. The (x, y) coordinates are given in meters. Calculate the resultant force experienced by the charge at the origin. Answer Firstly, sketch a schematic to represent the situation. We have three points, (0, 0), (0.3, 0) and (0.2, 0.2) specified on an xy axis, with distances in metres. y 0.2m 2µC 0.2m We are asked to calculate the force on the charge at the origin, and although we as yet do not know the magnitudes, we can mark the direction of the individual forces. FA is the force of the 2µC charge on the charge at the origin, and since the two charges are opposite in sign, this force is attractive, acting along the line between the two charges towards the 2µC charge. FB is repulsive, and therefore acts along the line between the two charges but points away from the 2µC charge as shown. 3µC y 45 45 FA 0.3m 2µC x x We can now calculate the magnitudes of FA and FB from Coulomb s Law. This is easiest for FA since r = 0.3m: FB FF AA = qq 1qq 2 4ππεε 0 rr 2 = (3 10 6 CC)( 2 10 6 CC) 4 3.14 8.85 10 12 0.3 2 = 0.6NN 7
For FB, r is the dashed line and by Pythagoras, rr = 0.2 2 0.2 2 = 0.28mm. Then; FF BB = qq 1qq 2 4ππεε 0 rr 2 = (3 10 6 CC)(2 10 6 CC) 4 3.14 8.85 10 12 0.28 2 = 0.69NN We can now calculate the resultant R on the charge at the origin; RR xx = FF AAAA FF BBBB = 0.6NN 0.69cccccc45 = 0.11NN y RR yy = FF AAAA FF BBBB = 0 0.69ssssss45 = 0.49NN Then FA 45 x And using RR = RR xx RR yy = 0.11 2 0.49 2 = 0.48NN θθ = ttttnn 1 RR yy RR xx The angle at which the resultant acts is or 77.3 below the horizontal. θθ = ttttnn 1 0.49 = 77.3 0.11 R FB The Coulombic repulsion between charges gives rise to a certain static distribution of charge over surface. Consider a solid metal sphere that is negatively charged and therefore has an excess of electrons. The free electrons on the sphere are all negative point charges, and each charge exerts a repulsive force on all other electrons on the sphere. The closer the electrons are on the sphere the greater the force of repulsion. To minimize energy, the electrons distribute themselves due to this force, such that the maximum distance between nearest neighbors is achieved. If you like, you can conceptualise it by remembering that the electrons don t like each other, and want to be as far away from each other as possible. The electrons therefore arrange themselves evenly over the surface, not throughout the volume, as shown by Figure G.1.2.2, maximising the distance r. If the surface were a metal plate, the electrons would again distribute themselves evenly over the surface, with a common distance between each electron and its nearest neighbors. Figure G.1.1.2 8
Exercise G.1.2 Assume p = 1.6 10 19 C, e = 1.6 10 19 C, ε0 = 8.85 10 12 C 2 /Nm 2 and π = 3.14 1. State in each case whether the force experienced between the two particles is attractive or repulsive. A. Proton and a proton B. p and e C. q1 and q2 D. q2 and q1 2. Two electrons are separated by a distance of 5cm. Calculate the Coulombic force between them, stating whether the force experienced is attractive or repulsive. 3. A hydrogen atom consists of a single electron orbiting a proton. The mean radius of the hydrogen atom (the Bohr radius) is 5.29 10 11 m. Calculate the Coulombic force of attraction exerted by the proton on the electron. 4. If the force between two charges, q1 = 2μC and q2 = 4μC is 0.5N, calculate the separation between the charges. Without further calculation, state the force between the charges if this separation distance is halved. Challenge Question 5. The following three charges are situated along a straight line (the positive x axis) with the charge and separations shown. Calculate the resultant force on charge q2, stating the direction of the resultant. q3 = 3µC q3 = 4µC q3 = 2µC x 3cm 1.5cm 9
G.1.3. Electric Field Electromagnetic force, just like the force of gravity, acts at a distance. There does not need to be any contact between the moon and the Earth for the moon to experience gravity, and there does not have to be any contact between two or more charges for them to feel an electromagnetic force of attraction or repulsion. The electric field E is a concept that describes the region of space around a charge, specifically it describes how much force another charge would experience at that position in space. The electric force experienced by a charge is exerted on it by the electric field created by other charges. The electric field that a point charge creates is radial, since it emanates from a single point in space, as shown in Figure G.1.3.1. The electric field is a vector and has size and direction, with the direction signified by arrows. Electric field ALWAYS starts on a positive charge and ends on a negative charge _ And the electric field strength is defined as follows; Figure G.1.3.1 The electric field strength E at a point in the field is defined as the force F per unit charge Q on a positive test charge placed at that point or in symbols EE = FF QQ The unit of E is the Newton per Coulomb (NC 1 ). F Consider Figure G.1.3.2. If a positive charge Q is placed in an electric field it experiences a force in the direction of the field line, and accelerates along the field line. If a negative charge is placed at that point, it experiences a force in the opposite direction to the field lines, and accelerated along the field line in the opposite direction to the field line. This is commensurate with Coulomb s Law, since there is always a positive charge at the beginning of a field line, and the charge is accelerated away from it if its positive, or towards it if it is negative. The closer the field lines are together, the greater E F x y Figure G.1.3.2 10
the force experienced. A charge placed at point x will experience a greater force than an identical charge at point y. Worked Example. An electron e is placed in a field due to a 2μC charge and experiences a force of 2 10 14 N. Sketch the field lines from the 2μC charge and the direction of the force experienced by the electron. Calculate the strength of the field at the electrons position. Answer The 2μC charge is positive and therefore the field points away from the charge. The electron is negative and therefore experiences a force along the field line in the opposite direction, toward the 2μC charge. We can choose an arbitrary position for the test charge (the electron), since we will then define the field strength at that point. 2 10 14 N 2µC e At a point in the field, the electron experiences a force of 2 10 14 N. Since E = F/Q, EE = FF ee = 2 10 14 NN 1.6 10 19 CC = 125000NNCC 1 Note that it is the force experienced by the electron, and therefore the charge of the electron, not the 2μC charge that is used to determine the electric field strength at a point in the field. So far we have considered the electric field from isolated static charges. If there are multiple static charges the resulting field is a combination of the two charges, as shown in Figure G.1.3.3. Recall that field lines always start on positive charges and always end on negative charges. Therefore when the two charges are of opposite sign the field lines go from the positive charge to the negative charge as shown by Figure G.1.3.3. (a). When the charges are of the same sign the field lines are deflected. _ (b) (a) Figure G.1.3.3 11
Exercise 1.3. 1. An electron is placed in the radial electric field of a positive charge. As a result the electron experiences a force and accelerates. Does the electrons acceleration A. Remain constant with time B. Decrease with time C. Increase with time Explain your answer. 2. Sketch the electric field distribution between two identical positive point charges separated by a distance r. Then sketch the distribution if one charge was twice the magnitude of the other. 3. A test charge of 2μC is placed in an electric field E = 120NC 1. Calculate the force experienced by the test charge. 4. A test charge of 4nC experiences a force of 0.03mN when placed in an electric field. Calculate the field strength at that point. Challenge Question 5. A test charge placed in an electric field experiences an acceleration a = 18ms 2. If the mass of the charge is 2 10 10 kg, calculate the field strength at that point. 12