1.2 Energy of Charged Particles

Transcription

1 1.2 Energy of Charged Particles Objective 1: Recall Coulomb s Law which states that the electrostatic force between 2 charged particles is inversely proportional to the square of the distance between them. Objective 2: Recall that an electric field originates at a positive charge and terminates at a negative charge, and that an electric field is an example of a vector field. Objective 3: Relate the electrostatic force between 2 charged particles to the electric field surrounding one or multiple charges. Objective 4: Understand the change in potential energy associated with the movement of a charge through an electric field. Objective 5: Understand the law of conservation of energy and how potential energy of a charged particle can be transformed into kinetic energy. Objective 6: Calculate the velocity of an electron accelerated as a result of a potential difference between 2 oppositely charged plates.

2 1.2.1 Coulomb s Law (Force) Objective 1: Recall Coulomb s Law which states that the electrostatic force between 2 charged particles is inversely proportional to the square of the distance between them. Two charges with similar charge will experience a force of repulsion (repel each other), and two charges with opposite charge will experience a force of attraction. The force between 2 charges (q 1 and q 2 ) is inversely proportional to the square of the distance (r) between them, and can be modeled with Coulomb s Law, 4p where is the permittivity of free space equal to 8.85 x F/cm (1F = 1C/V). If q 1 has the same sign as q 2, then the force between them is positive and they will repel each other. If q 1 and q 2 have opposite signs, then the force between them is negative and the charges will be attracted to each other (see Fig. 1.3 below). (a) (b) q 1 q 2 q 1 q 2 Figure 1.3 Illustration of force between (a) 2 similar charges and (b) 2 opposite charges. Example 1.5: What is the force between two opposite charges like those shown in Fig. 1.3(b) if q 1, = 1.6 x C and q 2 = x C, if q 1 is 1 µm from q 2? Solution: From Coulomb s Law, the force between the 2 charges is an attractive force and is equal to, 4p The force that exists between the 2 charges is a result of the electric field created between the 2 charges (see Sections and 1.2.3). The positive charge experiences a force to the right and the negative charge experiences a force to the left. We shall see in Section that the resulting electric field between these two charges is to the right, from the positive charge to the negative charge.

3 1.2.2 Electric Field Lines Objective 1: Recall that an electric field originates at a positive charge and terminates at a negative charge, and that an electric field is an example of a vector field. The electric field always originates at a positive charge and terminates at a negative charge as shown below. Figure Electric field between 2 point charges. Source: 20Physics/Fields/Images%20400/img_tb_4734.gif If the charges are sufficiently away from each other, the charge can be modeled as an isolated charge, Figure 1.5 Electric Field lines from a point charge assuming other charges are far removed from point charge. Source: Based on Coulomb s Law, the force exerted by a charge varies with distance away from any one charge. As a result the magnitude and direction of an electric field vary for each point in space. For each point in space, an electric field has an infinite set of vectors instead of a single point vector. This is called a vector field, and an electric field is one example of a vector field. Other examples include the velocity of fluid motion. For any given point in a fluid, there is a velocity vector in many different directions with different magnitudes. A uniform electric field can be created between 2 charged plates as shown in Fig. 1.6.

4 Figure 1.6 Uniform electric field between two parallel charged plates. Source: Figure 1.6 illustrates an electric field with a constant magnitude and direction, where the field lines are parallel and uniformly spaced. A high electric field has closely spaced lines, whereas a low electric field has widely spaced lines. A charge placed in a uniform electric field will experience a force resulting in the movement of the charge with a velocity that is proportional to the force exerted by the field (described in section 1.2.5). Similarly, the movement of charge in a conductor means that an electric field exists in the conductor. The force resulting from the electric field moves the electrons in a direction opposite to the electric field, and current is said to move in the direction of the electric field. If there is no current in a conductor, we know that the electric field in the conductor is zero.

5 1.2.3 Electric Field Objective 1: Relate the electrostatic force between 2 charged particles to the electric field surrounding one or multiple charges. An electric field exists on a point charge in space (point P) due to an electrostatic force from a positive charge (point P ). The charge at point P is assumed to be very small and to have no effect on the electric field (i.e. very small force). P P The electric field at point P is equal to the force (F) between the two charges divided by the charge (q) at point P (the point where the electric field is being measured), where the charges are both assumed to be positive in this case. If the point charge at P is moved further away from P, the force exerted by the positive charge on the charge at P will decrease, and consequently the electric field experienced by the charge at P will decrease. When one of the charges is negative, the electric field becomes, The units for E are volts per unit length (usually expressed as V/m or V/cm). (In the above equation, both F and E are vectors, and q is a scalar quantity.) From Coulomb s Law, the force on a test charge q 1, a distance r from a second charge q 2 is equal to, 4p The magnitude of the electric field at the point where q 1 is located is equal to,

6 Substituting the equation for force (F) into the equation for Electric Field (E) results in, 4p 4p If both charges are positive, the charge q 1 will experience a force to the left, and the electric field will be in this direction since the charge is positive and positive charges move in the direction of the electric field. q 1 q 2 The electric field at q 2 is said to be positive, or from left to right. If the charges have opposite charges, then the charges will experience an attractive force and The electric field is oriented in a direction opposite to the motion of the negative charge, or from the positive charge to the negative charge. Thus, the motion of an electron in an electric field is always opposite to the direction of the electric field. q 1 q 2 E - In order to determine the electric field at a given point from multiple charges, the following expression can be used, 4p

7 Example 1.6: What is the electric field on a positive charge q 1, if it is located 30 cm from a charge q 2, where q 2 = x C? Solution: The magnitude of the electric field at the location of q 1 is equal to, or 4p 4p The electric field at q 1 is equal to, Therefore, there is a force on charge q 1 towards q 2 and the direction of the electric field is from the positive charge to the negative charge. Note that the magnitude of the electric field is independent of the charge located at the point where the E field is being measured. Example 1.7: What is the electric field on a positive charge q 1, if it is located half way between 2 negative charges (q 2 = x C and q 3 = x C ) that are spaced 10 cm apart? - - q2 q1 q3 Solution: The magnitude of the electric field on the charge q 1 is equal to, 4p 4p

8 Therefore, there is a force on charge q 1 as a result of the negative charge at q 2 pointing towards q 2 and a second force on q 1 as a result of the negative charge at q 3 pointing towards q 3. Therefore, the electric field resulting from q 2 is negative and the electric field resulting from q 3 is positive. Because q 1 is placed midway between q 2 and q 3, and because q 3 > q 2, the electric field resulting from q 3 will be greater than the electric field from q 2. This results in a positive electric field on q 1 towards q 3.

9 1.2.4 Potential Energy Objective 1: Understand the change in potential energy associated with the movement of a charge through an electric field. When a charged particle q 1 moves in an electric field (caused by a charge q 2 a certain distance away), the electric field does work on the particle. The work done on the charged particle q 1 is equal to the change in its potential energy as the particle moves from distance r 1 to r 2 away from the charge q 2. The positive charged particle moves in the direction of the electric field, and its potential energy decreases in this direction, where is less than. q 2 E q 1 q 1 E L r 1, U 1 r 2, U 2 Figure 1.7 Potential energy change of a charged particle as it moves in an electric field between 2 charged particles separated by a distance L. The electric field surrounding charge q 2 will exert a force on the charge q 1 located at r 1, moving the charge from r 1 to r 2, away from q 2, resulting in a decrease in the potential energy of q 1. The magnitude of the force on q 1 is given by coulomb s law, and the change in the potential energy required to move q 1 from a distance r 1 to a distance r 2 is equal to the work done by the electric field where, The negative sign indicates that the potential energy is decreasing in this direction. 4p 4p

10 p 1 4p 1 Since, then the potential energy when q 1 is at r 1 is equal to 4 p and the potential energy of q 1 when it is at r 2 is equal to 4 p Therefore, U for multiple charges creating an electric field on a charge q at any given location along the path from 1 2 is equal to 4 p r 2 r 1 The work depends only on the end points and the work is the same for all possible paths from 1 to 2.

11 Example 1.8: A particle with a mass of 5 g and a charge of 2 x 10-9 C (q) starts from rest at point 1 and moves in a straight line to point 2 as shown in the figure below. What is the change in potential energy for this particle? q 1 =3 x 10-9 C 1 2 q 2 = -3 x 10-9 C 1 cm 1 cm 1 cm Solution: U q 4πε q r q r 3 10 C 1cm 3 10 C U 2 x 10 C 4p cm F J V C C cm J q r 3 10 C 2cm 3 10 C 1cm U 2 x 10 C 4p cm F J J C cm V C C cm

12 Therefore, the particle experiences a decrease in potential energy (U 1 > U 2 ) as it moves from point 1 to point 2 away from q 1.

13 1.2.5 Kinetic Energy Objective 5: Understand the law of conservation of energy and how potential energy of a charged particle can be transformed into kinetic energy. Objective 6: Calculate the velocity of an electron accelerated as a result of a potential difference between 2 oppositely charged plates. As mentioned in section 1.2.4, when a positively charged particle responds to an electric field, it moves in the direction of the electric field, and its potential energy decreases as the particle moves in this direction. Similarly, when a negatively charged particle responds to an electric field, it moves in the direction opposite that of the electric field, and its potential energy decreases as the particle moves in this direction. This decrease in potential energy corresponds to an increase in the kinetic energy of the particle based on the conservation of energy laws. This is similar to a ball being thrown up in the air, or an oscillating pendulum. As the ball moves up, it slows down, resulting in a decrease in kinetic energy. At its maximum height, it has maximum potential energy and v = 0 thus its kinetic energy is zero. When it begins to fall back to Earth, its kinetic energy increases until it reaches the ground, at which point its potential energy is zero. An oscillating pendulum swings from left to right, reaching a maximum height, before it swings back down to a minimum height, and then back to a maximum height on the opposite side. When the pendulum reaches a maximum height, its velocity is zero and its potential energy has a maximum value. As it swings down, it reaches it minimum height, at which point it has maximum velocity and its kinetic energy is maximum. Its kinetic energy is equal to its potential energy half way between its minimum and maximum points. In both cases, kinetic energy is being converted to potential energy and vice versa. Based on the law of conservation (discussed in section 1.1.1), as a particle moves along a path from an initial location (1) to a final location (2), the exchange in energy, between kinetic and potential must be equal or,

14 Example 1.9: For the charged particle described in Example 1.8, determine the velocity of the particle when it reaches point 2, if its initial velocity at point 1 was zero? Solution: If we assume that the loss in potential energy is converted to kinetic energy, then 1 2 and since the velocity at point 1 is zero, then K 1 is zero and In order to determine if relativistic calculations are necessary, it is required to calculate mc 2 for the 5 g charged particle. If the K.E. is greater or equal to 0.01 mc 2, then relativistic calculations are required, otherwise, classical equations may be used. The value of 0.01 mc 2 is equal to Since the kinetic energy of the particle is much less than this value, it is possible to use the classical equation for kinetic energy to obtain an accurate value for the velocity. Since

15 1.2.6 Potential (V) and Potential Energy (U) Potential is defined as potential energy per unit charge, 1 A potential exists at every point in an electrostatic field. Figure 1.8 illustrates a uniform electric field between 2 oppositely charged plates separated by a distance L, with equally spaced field lines from the positively charged plate to the negatively charged plate. E a x b Figure 1.8 Potential difference associated with charged particle at points a, x and b as a result of uniform electric field between 2 parallel charged plates separated a distance L. The potential difference between points a and x is equal to L and solving for The equation for V x is valid for a uniform electric field only, where the potential decreases linearly with x. At point b, where x L and Vx Vb, Solving for E,

16 Consequently, the electric field is equal to the potential difference between the plates divided by the distance between them. Example 1.10: When the terminals of a 100 V battery are connected to 2 large parallel plates 1 cm apart, what is the value of the uniform electric field between the plates? Solution: Since the electric field is equal to the potential difference between the plates divided by the distance between the plates, /

17 Example 1.11: The direction of a 100 V/cm electric field is in the vertical direction opposite to the gravitational field. Calculate the force on an electron in the electric field. Compare this force to the weight of the electron. Recall that m = 9.11 x kg and g = 9.8 m/s 2. Solution: Since, and The force on the electron due to gravity is equal to The ratio of the electrical to the gravitational force is Thus, the gravitational force is negligibly small compared to the electric force.

18 Example 1.12: The speed of an electron having a wavelength of m (the order of magnitude of atomic spacing in crystals) is 7.27 x 10 6 m/s. Through what potential difference must an electron be accelerated to acquire this speed? Solution: If we assume that the electron was initially at rest, then K 1 = 0 and U = K 2. Therefore, we need to calculate the kinetic energy of the electron with v = 7.27 x 10 6 m/s. Do we use the classical or the relativistic equation for kinetic energy? We can use the classical equation for kinetic energy if v << 0.1 c. The velocity in this example is less than 0.1c or 3 x 10 7 m/s. Furthermore, v/c =. / / = 0.02, which proves that v <<0.1c. Consequently, we can use the classical equation for kinetic energy which is equal to the change in potential energy of the particle ΔV ΔV 150V Note the resulting unit analysis if using units of ev for energy instead of Joules, ΔV

19 Therefore, the change in potential in volts is numerically equal to the change in potential energy in units of ev. Since and the potential energy of a charge q in an electric field caused by 1 or more charges is equal to U q q 4πε r then the potential V at any point in the electric field is equal to V U q 1 q q 4πε q r or V 1 4πε q r

20 References 1. F.W. Sears, Zemansky, Young, Addison Wesley Education Publishers, 1991.