Lesson 1: Electric Force and Electric Field PDF

Outlines 1. Coulomb’s Law 2. Electric field due to a point charge 3. Electric Field of a Continuous Charge Distribution 4. Electric Field Lines 5. Motion of a Charged Particle in a Uniform Electric Field 1. Coulomb’s Law  There are two kinds of the electric charges in the nature, which are positiv...

Outlines 1. Coulomb’s Law 2. Electric field due to a point charge 3. Electric Field of a Continuous Charge Distribution 4. Electric Field Lines 5. Motion of a Charged Particle in a Uniform Electric Field 1. Coulomb’s Law  There are two kinds of the electric charges in the nature, which are positive and negative charges.  The charges of opposite sign attract one another and the charges of the same sign repel one another.  The Charge is quantized  The charge of an isolated system is conserved, which means that charge cannot be created nor destroyed, it can only be transferred from one system to another Units of the Electric Charge 1 milli coulomb (1 mC) = 10-3 C 1 micro coulomb (1 µC) = 10-6 C 1 nano coulomb (1 nC) = 10-9 C 1 pico coulomb (1pC) = 10-12 C Charles Coulomb (1736–1806) measured the magnitudes of the electric forces between charged objects and he found that:- The electric force is inversely proportional to the square of the separation r between the 1 particles and directed along the line joining them F ∝ ; r2 The electric force is proportional to the product of the charges q 1 and q 2 on the two particles F ∝ q1 q 2 ; 1  The electric force is attractive if the charges are of opposite sign and repulsive if the charges have the same sign; The electric force is a conservative force. We can express Coulomb’s law as an equation giving the magnitude of the electric force (sometimes called the Coulomb force) between two point charges: q1 q 2 Fe = K e r2 Where k e is a constant called the Coulomb constant. The value of the Coulomb constant depends on the choice of units. The Coulomb constant k e in SI units has the value: 1 K= = 9 × 10 9 N.m 2 /C 2 4πε 0 Where the constant ε o (lowercase Greek epsilon) is known as the permittivity of free space and has the value ε o = 8.85 10 -12 C 2 / N.m 2 The smallest unit of charge e known in nature is the charge on an electron (- e) or a proton (+ e) and has a magnitude e = 1.602 x 10 −19 C Example. 1: Forces in hydrogen atom The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately 5.29 x 10-11 m. Find the magnitudes of the electric force and the gravitational force between the two particles. 2  Figure. 2 Solution From Coulomb’s law, we find that the magnitude of the electric force is Using Newton’s law of universal gravitation and above Table for the particle masses, we find that the magnitude of the gravitational force is: FG = G mM r2 FE 8.23 × 10 −8 = = 2.27 × 10 39 FG 3.62 × 10 −47 Thus, the gravitational force between charged atomic particles is negligible when compared with the electric force. Figure. 3 When dealing with Coulomb’s law, you must remember that force is a vector quantity and must be treated accordingly. The law expressed in vector form for the electric force exerted by a charge q 1 on a second charge q 2 , written F 12 , is: q1 q 2 ∧ F12 = K e r r2 Where ȓ is a unit vector directed from q 1 toward q 2 , as shown in Figure2-a. Because the electric force obeys Newton’s third law, the electric force exerted by q 2 on q 1 is equal in magnitude to the force exerted by q 1 on q 2 and in the opposite direction; that is, F 21 = - F 12 3  If five charges are present, as shown in figure. 4, then the resultant force exerted by particles 2, 3, 4, and 5 on particle 1 is: F1 = F21 + F31 + F41 + F51 Figure. 4 Example. 2: Find the resultant force exerted on Q1 = 1 μC charge? Figure. 5 Solution First, note the direction of the individual forces exerted by Q 2 and Q 3 on Q 1. The force F 21 exerted by Q 2 on Q 1 (separated by distance 3cm) is attractive becauseQ 2 and Q 1 have opposite signs and. The force F 31 exerted by Q 3 on Q 1 (separated by distance 3cm) is repulsive because both charges are positive. The magnitude of F 21 is: 4  The magnitude of F 31 is: We can also express the resultant force acting on Q 1 in unit vector form as 2. Electric field due to a point charge The electric field vector E at a point in space is defined as the electric force F e acting on a positive test charge q 0 placed at that point divided by the test charge: Fe E = qo The vector E has the SI units of newtons per coulomb (N/C). Note that: E is the field produced by some charge or charge distribution separate from the test charge; it is not the field produced by the test charge itself. Also, note that the existence of an electric field is a property of its source; the presence of the test charge is not necessary for the field to exist. The test charge serves as a detector of the electric field. The Equation electric field can be rearranged as: Fe = E qo Where we have used the general symbol q for a charge. This equation gives us the force on a charged particle placed in an electric field. If q is positive, the force is in the same direction as the field. If q is negative, the force and the field are in opposite directions ( as shown in the following figure). 5  Figure. 1 To determine the direction of an electric field, consider a point charge q as a source charge. This charge creates an electric field at all points in space surrounding it. A test charge q 0 is placed at point P, a distance r from the source charge, as in Figure 1-a. According to Coulomb’s law, the force exerted by q on the test charge is: q qo ∧ Fe = K e r r2 Fe  E= qo q ∧ ∴ E = Ke r r2 Where ȓ is a unit vector directed from q toward q 0. If the source charge q is positive, Figure 1-b shows the situation with the test charge removed; the source charge sets up an electric field at point P, directed away from q. If q is negative, as in Figure 1-c, the force on the test charge is toward the source charge, so the electric field at P is directed toward the source charge, as in Figure 1-d. Note that at any point P, the total electric field due to a group of source charges equals the vector sum of the electric fields of all the charges. Thus, the electric field at point P due to a group of source charges can be expressed as the vector sum 6  ∧ Where r i is the distance from the ith source charge q i to the point P and ri is a unit vector directed from q i toward P. Example. 3: Electric Field Due to Two Charges A charge q 1 = 7.0 µC is located at the origin, and a second charge q 2 = -5.0 µC is located on the x axis, 0.30 m from the origin (Figure 2). Find the electric field at the point P, which has coordinates (0, 0.40) m. Figure. 2 Solution First, let us find the magnitude of the electric field at P due to each charge. The fields E 1 due to the 7.0 µC charge and E 2 due to the - 5.0 µC charge are shown in Figure 2. Their magnitudes are: 7 The vector E 1 has only a y component. The vector E 2 has an x component given by E 2 3 4 cos θ = E 2 and a negative y component given by E 2 sin θ = E 2. Hence, we can 5 5 express the vectors as The resultant field E at P is the superposition of E 1 and E 2 : From this result, we find that E makes an angle φ of 66° with the positive x axis and has a magnitude of 2.7 x 105 N/C 3. Electric Field of a Continuous Charge Distribution Figure. 4 To evaluate the electric field created by a continuous charge distribution, we use the following procedure: first, we divide the charge distribution into small elements, each of which contains a small charge 0q, as shown in Figure 4. Next, we use the following Equation to calculate the electric field due to one of these elements at a point P. Finally, we evaluate the total electric field at P due to the charge distribution by summing the contributions of all the charge elements. The electric field at P due to one charge element carrying charge ∆q is where r is the distance from the charge element to point P. The total electric field at P due to all elements in the charge distribution is approximately 8 Where the index i refers to the ith element in the distribution. Because the charge distribution is modeled as continuous, the total field at P in the limit ∆q i → 0 is Where the integration is over the entire charge distribution. We illustrate this type of calculation with several examples, in which we assume the charge is uniformly distributed on a line, on a surface, or throughout a volume. When performing such calculations, it is convenient to use the concept of a charge density along with the following notations: 3.1 Volume charge density If a charge Q is uniformly distributed throughout a volume V, the volume charge density is ρ defined by: Where ρ has units of coulombs per cubic meter (C/m3). 3.2 Surface charge density If a charge Q is uniformly distributed on a surface of area A, the surface charge density σ is defined by: Where σ has units of coulombs per square meter (C/m2). 3.3 Linear charge density If a charge Q is uniformly distributed along a line of length L , the linear charge density λ is defined by: Where λ has units of coulombs per meter (C/m). If the charge is nonuniformly distributed over a volume, surface, or line, the amounts of charge dq in a small volume, surface, or length element are Example. 4: The Electric Field Due to a Charged Rod 9 A rod of length L has a uniform positive charge per unit length λ and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end (Figure 5). Figure. 5 Solution Let us assume that the rod is lying along the x axis, that dx is the length of one small segment, and that dq is the charge on that segment. Because the rod has a charge per unit length λ, the charge dq on the small segment is dq = λ dx. The field dE at P due to this segment is in the negative x direction (because the source of the field carries a positive charge), and its magnitude is a+L a+L dq λ dx E= ∫a K 2 = x ∫a K x2 Where the limits on the integral extend from one end of the rod (x = a) to the other (x=L+ a). The constants k e and λ can be removed from the integral to yield a+L a+L dx  1 E = Kλ ∫ a x 2 = Kλ  −   xa 1 1   L  E = Kλ  −  = Kλ   a a + L  (a + L )a  Example. 5: The Electric Field of a Uniform Ring of Charge A ring of radius a carries a uniformly distributed positive total charge Q. Calculate the electric field due to the ring at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring (Figure 6 - a). 10  Figure. 6 Solution The magnitude of the electric field at P due to the segment of charge dq is dq dE = K e r2 This field has an x component dE x = dE Cos θ along the x axis and a component dE ⊥ perpendicular to the x axis. As we see in Figure 6-b, however, the resultant field at P must lay along the x axis because the perpendicular components of all the various charge segments sum to zero. That is, the perpendicular component of the field created by any charge element is canceled by the perpendicular component created by an element on the opposite side of the ring. Because r = x2 + a2 and Cos θ = x/r, we find that dq dq dE = K =K 2 r 2 a + x2 dE x = dE Cos θ  dq  x  E x = ∫ dE cos θ = ∫  K e 2  2   a + x2 (  a + x 2 )12   Ke x = (a 2 + x2 ) 12 ∫ dq q x E = Ke (a 2 + x2 ) 32 This result shows that the field is zero at x = 0. Does this finding surprise you? Example.6: The Electric Field of a Uniformly Charged Disk 11 A disk of radius R has a uniform surface charge density σ. Calculate the electric field at a point P that lies along the central perpendicular axis of the disk and a distance x from the center of the disk (Figure 7). Figure. 7 Solution If we consider the disk as a set of concentric rings, we can use our result from Example 4 - which gives the field created by a ring of radius a and sum the contributions of all rings making up the disk. By symmetry, the field at an axial point must be along the central axis. The ring of radius r and width dr shown in Figure 7 has a surface area equal to 2 π r dr. The charge dq on this ring is equal to the area of the ring multiplied by the surface charge density: dq = 2 π σ r dr. Using this result in the equation given for Ex in Example 4 (with a replaced by r), we have for the field due to the ring x dE x = K e (2 π σ r dr ) (x 2 + r2 ) 32 To obtain the total field at P, we integrate this expression over the limits r = 0 to r = R, noting that x is a constant. 12  R 2rdr Ex = Ke x π σ ∫ (x 0 2 + r2 ) 32 ∫ (x ) ( ) R −3 2 = Ke 2 π x σ 2 + r2 d r2 0 ∫ (x ) ( ) R 1 −3 2 = 2 π x σ 2 + r2 d r2 2 π εo 0  xσ  ∫ (x ) ( ) R −3 2 =   2 + r2 d r2  4ε 0  0 ( ) R  x σ   x2 + r 2 −1 2  =      4 ε0   −1 2  0     Then, E x =  σ  1 − x   2ε 0   (x 2 +R ) 2 12  When x→ 0 or R → ∞ σ Ex = 2 ε0 4. Electric Field Lines The rules for drawing electric field lines are as follows: The lines must begin on a positive charge and terminate on a negative charge. In the case of an excess of one type of charge, some lines will begin or end infinitely far away. Figure. 8 The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge. 13  Figure. 9 No two field lines can cross. Figure. 10 5. Motion of Charged Particles in a Uniform Electric Field When a particle of charge q and mass m is placed in an electric field E, the electric force exerted on the charge is qE. If this is the only force exerted on the particle, it must be the net force and causes the particle to accelerate according to Newton’s second law. Thus, Fe = q E = m a q E The acceleration of the particle is therefore a = m If E is uniform (that is, constant in magnitude and direction), then the acceleration a is constant. If the particle has a positive charge, its acceleration is in the direction of the electric field. If the particle has a negative charge, its acceleration is in the direction opposite the electric field. Example. 7: An Accelerating Positive Charge A positive point charge q of mass m is released from rest in a uniform electric field E directed along the x axis, as shown in Figure 11. Describe its motion. 14  Figure. 11 Solution The acceleration is constant and is given by: a = q E m The motion is simple linear motion along the x axis. Therefore, we can apply the equations of kinematics in one dimension: Choosing the initial position of the charge as x i = 0 and assigning v i = 0 because the particle starts from rest, the position of the particle as a function of time is: The speed of the particle is given by: The third kinematic equation gives us: from which we can find the kinetic energy of the charge after it has moved a distance: 15