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CHAPTER 23 ELECTRIC FIELD ELECTRIC CHARGE There are two kinds of electric charges: positive(protons) and negative(electrons). The negative charge of an electron and the positive charge of a proton have the same magnitude given as e 1.6 10 19 C Thus, an atom is electrically neutral when the number of electrons equals the number of protons. It is found, experimentally, that like charges repel each other and unlike charges attract each other. Charges have the following properties: Electric charge is conserved. Electric charge is quantized, i.e., q=ne With n=0,1,2, and e is the charge of the electron (proton). 2 CONDUCTORS AND INSULATORS A conductor is the substance that permits electric charge to move freely, e.g, Silver, copper, and aluminum An insulator is the substance that does not permit electric charges to move. e.g., glass, plastic, and rubber There are substances that fall in an intermediate position between conductors and insulators. These substances are called semiconductors such as silicon and germanium. With the properties of conductors and insulators in mind, one can charge a conductor by a technique called charging by induction, (see the Figure 1.1). he center of the circle. (a) (b) (c) Figure 1.1 Charging by induction. (a) The charge on a conducting sphere is redistributed when a charged rubber rod is brought near the sphere. (b) The access negative charges leave the sphere as it is grounded. (c) The rubber and the ground connection are removed, and the sphere has a uniform positive charge. 3 COULOMB’S LAW It is found that: (i) The force is directly proportional to the magnitude of the charges of each particle. (ii) The force is inversely proportional to the square of the distance between the two particles. (iii) The direction of the force is along the line joining the two particles. (iv) The force is attractive if the charges are of opposite signs and repulsive if the charges have the same signs. From these observations the magnitude of the force acting on a point charge q1 due to another point charge q2 can be expressed as F21 k q1 q2 2 r21 where r21 is the distance between the two charges, and k is the proportionality constant known as Coulomb’s constant. The direction of the force is determined according to the sign of the two charges. It is repulsion if the two charges are alike and attractive if they are unlike. The proportionality constant, k, has a numerical value depends on the choice of units. The SI unit of charge is Coulomb (C), which is considered as the fourth primary unit beside meter, kilogram, and second. In this system k has the value k 9.0 109 N.m2/C2 Coulomb’s constant is usually written as k 1 4o with o 8.8542 1012 C2/N.m2, is called the permittivity of free space. 4 The cgs unit of charge is statcoulomb or electrostatic unit (esu). In this system, the constant k is defined to be unity with no units. In vector form Coulomb's law can be written as qq F21k 12 2 r̂21 r21 where r̂21 is a unit vector directed from q2 to q1. It should be noted that the electric force is of the same form as the gravitational force. As the gravitational force is conservative, we conclude that the electrostatic force is also conservative. q1 r̂21 q2 If there are more than two charges, the total force acting on one charge due to the others is the vector sum of the forces due to the individual charges. For example, if there are three charges, then the net force on particle 1 due to the other two particles is given as F1F21F31 Always remember that the electric force obeys Newton’s third law, that is F21 F12 Remark: Coulomb’s law applies only for point charges. 5 Example 23.1 Find the magnitude of the electrical force and the gravitational force between the electron and the proton in the H-atom. The average distance between the two particles is 0.53 Ao. Solution q1 q2 Fe k (9 10 ) 9 r122 (1.6 1019 )2 (0.53 10 10 2 ) 8.2 108 N To find the gravitational force we have Fg G m1m2 r122 6.67 101011 )(1.67 10 (9.11(10 0.53 10 ) 31 27 ) 10 2 3.6 10 47 N F Note that e 21039 Fg Example 23.2 Three point charges of 5 C, -4 C, and 8 C are located at the corners of an equilateral triangle of side 0.1 m as shown in the Figure. Find the resultant electric force exerted on q3. Solution Here the charge q3 are affected by two forces, F13 due to q1 and F23 due to q2. Now F13 k q1 q3 r132 F13 q2= - 2.0 C 0.1 m q1=5.0 C 0.1 m F23 q3=5.0 C 2 0.1 m 6 (9 10 ) 9 (5.0 106 )(5.0 106 ) ( 2 0.1)2 and F23 k (9 109 ) 11 N q2 q3 r132 (2.0 106 )(5.0 106 ) (2)2 9.0 N In vector notation we have F13 11cos iˆ 11sin ˆj 7.9 iˆ 7.9 ˆj N And F23 9.0 iˆ N Now the resultant force acting on q3 is the sum of the two forces, i.e., F3F13F23 7.9 9.0iˆ 7.9 ˆj 1.1iˆ 7.9 ˆj N Example 23.2 Three point charges, q1 = 15C at the origin, q2 = 6.0 C at x=2.0 m, and q3 at x. If the resultant electric force exerted on q3 is zero, find x. 2.0 m 2.0 - x x q1=15C F13 q3 F23 q2= 6.0 C Solution The two forces acting on q3 are both attractive and so opposite. For the net force to be zero their magnitudes must, therefore, be equal. i.e., F13 F23 k q1 q3 r132 k q2 q3 2 r23 7 k 15 10 6 2 x q3 2 k 6.0 10 x 6 q3 2 5 x 2 8 4 x 2 x 2 3x 2 8 x 8 0 x=0.78 m and x=-3.4 m The second solution is not acceptable since at a point to the left of q2 the two forces will be in the same direction, and so their sum can't be zero at all even if their magnitudes are equal. Example 23.4 Two identical small charged spheres, each of mass 30 g and charge q are hanged in equilibrium as shown in the Figure. The length of each string is 0.15 m, and the angle = 5.0o . Find q. Tcos 0.15 m 0.15 m T Fe Tsin 2a mg (a) (b) Solution It is enough to study the equilibrium state of one of the spheres. The forces acting on such a sphere is shown in the free-body diagram shown in Figure (b). The equilibrium conditions imply that Fx 0 and Fy 0 From the first condition we write Fe T sin (1) Here Fe is the electric force acting on the sphere by the other. The second condition gives 8 mg T cos (2) Dividing Equation (1) by Equation (2), we obtain tan Fe mg (3) Substituting the values of m, g, and in Equation (3) we get Fe 30 102 9.8 tan 5.0 2.6 102 N The magnitude of the electric force Fe is obtained from Coulomb’s law as Fe k q 2 r2 Solving for q and knowing that r 2a 20.15sin 0.013 m we get q Fe r 2 2.6 102 (0.013) 2 4.4 108 C 9 k 9 10 9 THE ELECTRIC FIELD Any charge has its own electric field within a region surrounding the charge. The electric field E at a point in space is a vector quantity defined as the electric force acting on a positive small test charge qo placed at that point divided by the magnitude of the test charge, i.e., F E qo The test charge should be small enough to ensure that its presence does not affect the charge distribution that produces E. Since qo is always positive, the direction of E is the direction of F. The SI unit of E is Newton per Coulomb (N/C). If a charge q is placed at a point with an electric field E, this charge experiences an electric force given by F qE ELECTRIC FIELD OF POINT CHARGES: Let us calculate the electric field at a point p arising from an isolated point charge q, a distance r from it. To do so we assume the existence of a test charge qo at point p, (see the figure). From Coulomb’s law the force upon the test charge is qq F k 2o rˆ r where the unit vector directed from q to the point p. Using Eq. 1.5 qo EP P r q 10 q E k 2 rˆ r It is clear from the definition of r̂ that if q is positive the direction of E is outward from the charge, if the charge is negative E it is directed toward q. The electric field at a point due to a group of point charges is the vector sum of the electric fields at that point due to each charge individually, i.e., if we have n charges the net electric field E is EE1E2 En where E1, E2, and En are the electric fields due to the charges q1, q2, and qn, respectively. Example 23.5 Two point charges of 7.0 C, -5.0 C, are located as shown in the figure. Find the resultant electric field at point P. E1 p E2 0.4 m 0.5 m q1=7.0 C 0.3 m q2= - 5.0 C 11 Solution a) Let us first calculate the magnitudes of the electric fields for each charge individually, E1 due to the 7.0-C charge and E2 due to the -5.0-C. E1k q1 r12 and E2 k q2 r22 9 109 9 10 9 7.(00.410) 6 3.9 105 2 5.0 106 (0.5) 2 1.8 106 N/C In vector notation we have E1 3.9 105 ˆj N/C And E2 1.8105 cos iˆ 1.8 105 sin ˆj 1.11059 iˆ 1.4 105 ˆj Now the resultant electric field at point P is E p E1E2 1.1105 iˆ 3.9 105 1.4 105 ˆj 1.1105 iˆ 2.5 105 ˆj N/C Example 23.6 A positive point charge q and a negative point charge –q are separated by a distance 2a (electric dipole). Find E at point P a distance y>>a from the origin. E+ Ey r q a a -q 12 Solution: The resultant electric field at P is E E E . Since P is equidistant from the two charges the magnitude of E+ and E- are equal and are given by E E k q r2 As it is clear from the directions of the two fields, the y-components of E+ and Ecancel each other. The net electric field at P is, therefore E E cos E cos i From the figure we see that cos d 2r , so we find E 2k Now r3 a2 y2 3 q r 2 cos k q2a r3 i 2 If ya we can neglect the first term in the bracket so that r3 is approximated to r 3 y3 . Ek q2a y 3 i=k p y3 where p q2ai is called the electric dipole moment. Its direction is taken to be from the negative charge to the positive charge. ELECTRIC FIELD OF A ONTINUOUS CHARGE DISTRIBUTION To evaluate the electric field of such a configuration, the following procedure are used: 1- We divide the distribution into small elements such that the electric field for each element can be calculated easily. 13 2- Each element is assumed a charge dq 3- Next, we calculate the electric field due to one of these elements at the point in question 4- Finally, we integrate over the charge distribution to evaluate the total electric field due to the whole charge distribution. Remark: The integration should be performed for each component of the electric field, why? Charge Density When dealing with continuous charge distribution it is convenient to use the concept of charge density. If the charge is distributed along a line we define the linear charge density as dq dl If the charge is distributed over a surface we define the surface charge density as dq dA If the charge is distributed within a volume we define the volume charge density as dq dV where dq is the charge of a small line, surface, or volume element. The SI units of the charge densities, , , and are C/m, C/m2, and C/m3, respectively. If the charge is uniformly distributed along a line, over a surface, or within a volume, the charge densities are constants. In this case the total charge Q can be obtained by integrating both sides of the above Equations Ql , QA , Or Q V 14 where l, A, and V are the length of the line, the area of the surface, and the magnitude of the volume, respectively. Example 23.7 A rod of length L lying along the x-axis is uniformly charged with a total charge Q. Calculate the electric field at a point p along the axis of the rod, a distance D from one end, as shown in the Figure. L D dE x dx dq Solution We divide the rod into small elements each of length dx and charge dq. Each element can be considered as a point charge, so the electric filed dE due to one of these elements a distance x from p is dEk dq x2 Note that each element produces a field in the negative x-axis as shown in Figure 1.-, and so we can integrate simply to get Ek dq x2 Expressing dq in term of the charge density as dq = dx we get L D E k D L D 1 k x x2 D dx k L D L D k Q DL D where we have substituted for L by Q. From this result we conclude that if p is far away from the rod (DL), then we recover the result of the point charge as expected, that is. 15 E k L D2 Example 23.8 A ring of radius R has a uniform charge distribution of magnitude Q. Calculate the electric field along the axis of the ring at a point p lying a distance D from the center of the ring. dq r R D x dE Solution: Again we divide the ring into small elements each charge dq. The e.f due to one element is dEk dq r 2 k R dq 2 D2 Because of the symmetry about the x-axis we should note that there exist a similar element, which is also located a distance r from p. This element produces an equal field dE but differing in direction. As shown in the figure, the vertical components of these two fields cancel one another and this will be hold for each such pair of elements. Hence, the total field is the sum of the x-components of dE which is given as dE x dE cos k R Ddq 2 D2 3 2 Integrating this expression we get 16 Ex k R D 2 D2 3 2 dq Ex k R DQ 2 D2 3 2 Again for the limiting case DR we recover the result of the point charge. At the center of the ring D=0 and E must be zero, comment. Example 23.9 A disk of radius R has a uniform charge density . Calculate the electric field along the axis of the disk at a point p lying a distance D from the center of the disk. Solution: In this problem we divide the disk into thin rings each of thickness dr. From the result of the previous example, the electric filed due to one element with radius r is dE k r Ddq 2 D2 3 R r p D dr 2 Integrating the last result and noting that dqdA 2 rdr we get R E kD 0 r 2rdr 2 D2 3 2 2 D r2 kD 12 12 R D 2k 1 D2 R2 0 The electric field for a large charged plane can be obtained from the last result by letting R (or x0). This gives E 2k 2 o 17 THE ELECTRIC FIELD LINES The electric filed in a region is represented by imaginary lines known as electric field lines (lines of force) introduced by Michael Faraday (1791-1867). These lines have the following properties: 1- The direction of the lines at any point is the direction of the electric field at that point. 2- The lines must begin on positive charges and terminate at negative charges. 3- The number of lines per unit area, perpendicular to E, is proportional to the magnitude of E in that region. 4- No two field lines can cross. Motion of Charged Particle in a Uniform E.F. The e. force is a force that obey Newton's laws, i.e., this force will creates an acceleration on the particle If a particle of mass m and charge q exists in a uniform e.f E the e. force acting on the charge is 18 Fe qE ma qE a m Since E is uniform a is constant and the formulas govern the motion with constant acceleration are still hold. Example 23.10 A positive point charge q of mass m is released from rest in a uniform e.f E E iˆ . Describe its motion. E v Solution The acceleration of the charge is qE qE ˆ a i m m Since vo 0 and the acceleration is along the x-axis the motion is linear. After a time t the particle is travelled a distance given by x vo t 12 at 2 qE 2 t 2m v vo at The speed of the charge at this time is Note that qE qE qE 2 t v 2 vo2 2ax 2 t v m m 2m Applying the conservation of energy principle Wnet K qEx 12 mv2 v 2qEx m qE 2 2qE t 2m qE t m m qE t m 19 Example 23.11 An electron is projected, horizontally with initial speed of 3.00 106 m/s into a region of uniform electric field between two plates of 200 N/C and directed vertically upward. The horizontal length of the plates is 0.1 m. a) Find the acceleration of the electron while it is in the field. b) How long is it in the field? c) What is the vertical displacement of the electron inside the field? ____________________ E y ++++++++++++++++++++++ l Solution: a) The acceleration of the electron is F qE a m m 1.6 1019 200 j 3.51 1013 j m/s2 31 9.11 10 This means that ax = 0. b) For the time we have t x 0.1 3.33 10 8 s. 6 v x 3.00 10 b) The vertical displacement can be calculated from the equation y v0 y t 12 a y t 2 12 3.511013 3.33 108 2 1.95 cm.