Ben Gurion University of the Negev www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter Physics 2B for Materials and Structural Engineering Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Irina Segel Week 3. Electric potential energy and potential – Conservative forces • conservation of energy • electric potential energy • potential energy of several charges • continuous charge distributions • electric potential Source: Halliday, Resnick and Krane, 5th Edition, Chap. 28. Conservative forces Suppose a force F acting on a particle depends only on the position r of the particle, i.e. F = F(r). The work W done by F in moving the particle from r1 to r2 is defined to be r2 W F(r) dr . r2 r1 If W does not depend on the choice of a path for r, then the force F is called conservative. Where does this name come from? r1 Conservative forces If F is conservative, then r2 r2 W F(r) dr F(r) dr , r1 r1 r2 where the color framing the integral indicates the path taken from r1 to r2. r1 Conservative forces If F is conservative, then r2 r2 0 F(r) dr F(r) dr , r1 r1 r2 where the color framing the integral indicates the path taken from r1 to r2. r1 Conservative forces If F is conservative, then r2 r1 0 F(r ) dr F(r ) dr r1 r2 r2 F(r ) dr . r1 Conservative forces Conclusion: if F is conservative, then r2 0 F(r ) dr r1 Conservative forces If F is conservative, then we can define a potential energy U(r) such that r2 U(r2) – U(r1) = F(r ) dr , r1 r2 and U(r) is a single-valued function of r (i.e. U(r) is a scalar field). With U(r) we can define a conserved total energy, the sum of the kinetic energy and the potential energy. r1 Conservation of energy If F is conservative, then we can define a potential energy U(r) such that r2 U(r2) – U(r1) = F(r ) dr , r1 r2 and U(r) is a single-valued function of r (i.e. U(r) is a scalar field). With U(r) we can define a conserved total energy, the sum of the kinetic energy and the potential energy. Conclusion: a conservative force implies a conserved total energy. r1 Conservation of energy Is the electric force a conservative force? Conservation of energy Is the electric force a conservative force? If not, then we can use an electric field to get energy for free. Here’s how: Conservation of energy Is the electric force a conservative force? If not, then we can use an electric field to get energy for free. Here’s how: If the electric force Fq on a particle of charge q is not conservative, then for some loop in r we can write 0 Fq (r ) dr . Fq Fq Fq Conservation of energy Is the electric force a conservative force? If not, then we can use an electric field to get energy for free. Here’s how: If the electric force Fq on a particle of charge q is not conservative, then for some loop in r we can write 0 Fq (r ) dr , Fq Fq which means that every time the charge makes a complete loop in r, it gets more kinetic energy! Fq Electric potential energy We will prove that the electrostatic force is conservative! First, we will calculate the work W done by the electric force if we move a charge q1 in a straight line towards a charge q2. Then we will prove that W does not depend on the path. Electric potential energy We will prove that the electrostatic force is conservative! First, we will calculate the work W done by the electric force if we move a charge q1 in a straight line towards a charge q2. Then we will prove that W does not depend on the path. The charge q2 is always at r2 while the charge q1 moves in a straight line to r1 from a point infinitely far away. r2 r1 Electric potential energy We will prove that the electrostatic force is conservative! First, we will calculate the work W done by the electric force if we move a charge q1 in a straight line towards a charge q2. Then we will prove that W does not depend on the path. The charge q2 is always at r2 while the charge q1 moves in a straight line to r1 from a point infinitely far away. r1 W r1 q1q2 F(r) dr 4 0 r2 r 2 dr 1 r q1q2 4 0 r2 r1 . F(r) dr r1 r r2 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? r2 r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? r2 F(r) dr r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? r2 F(r) dr r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? Since F(r) · dr is hard to compute… r2 F(r) dr r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? Let’s resolve dr into dr║ and dr┴: r2 F(r) dr r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? Let’s resolve dr into dr║ and dr┴: r2 F(r) dr r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? Resolved into dr║ and dr┴, the path looks like this… r2 F(r) r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? …which we arrange to look like this: r2 F(r) r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? And now we are back to the same integral we had before, since integration dr┴ does not contribute anything: r2 F(r) r r1 Electric potential energy But now suppose the charge q1 reached the point r1 by the path indicated. What is the work done by the electric force? And now we are back to the same integral we had before, since integration dr┴ does not contribute anything: r1 q1q2 W dr 2 4 0 4 0 r2 r1 r r 2 1 q1q2 F(r) r r1 . r2 Electric potential energy Conclusion 1: The electrostatic force is conservative. Electric potential energy Conclusion 1: The electrostatic force is conservative. Conclusion 2: The electric potential energy of two charges q1 and q2 at points r1 and r2, respectively, is q1q2 U 4 0 r2 r1 , regardless of how the charges were assembled. Remember Faraday? We had some basic rules and observations about field lines: •They never start or stop in empty space – they stop or start on a charge or extend to infinity. •They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. • The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. Remember Faraday? Now we can add a new basic rule about field lines: •They never start or stop in empty space – they stop or start on a charge or extend to infinity. •They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. • The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. • There are no closed loops! E E E E Remember Faraday? Now we can add a new basic rule about field lines: •They never start or stop in empty space – they stop or start on a charge or extend to infinity. •They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. • The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. • There are no closed loops! E E E E Electric potential energy What is the electric potential energy of three charges q1, q2 and q3 at points r1 r2 and r3, respectively? r3 r2 r1 Electric potential energy What is the electric potential energy of three charges q1, q2 and q3 at points r1 r2 and r3, respectively? It is q1q2 q1q3 q2 q3 U 4 0 r2 r1 r3 r1 r3 r2 1 r3 . r2 r1 Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) Fq θ F-q Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) dr θ Fq F-q Answer: U ( ) 2 Fq (r ) dr 2Fq (r ) dr Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) Not the same d! dr θ Fq F-q Answer: U ( ) 2 Fq (r ) dr 2Fq (r ) dr Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) dr θ Fq F-q Answer: U ( ) 2 Fq (r ) dr 2Fq (r ) dr Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) Δr θ F-q Fq Δr Answer: U ( ) 2 Fq (r ) dr 2Fq (r ) dr d 2( Eq) cos Eqd cos . 2 Electric potential We have seen that “electric potential energy” can include all the charge in a system – for example, we calculated the total potential energy of a system of three point charges. Or it can mean the potential energy of only some of the charges in an external electric field – for example, we calculated the potential energy of a dipole in a constant electric field. Electric potential We have seen that “electric potential energy” can include all the charge in a system – for example, we calculated the total potential energy of a system of three point charges. Or it can mean the potential energy of only some of the charges in an external electric field – for example, we calculated the potential energy of a dipole in a constant electric field. In the case of one point charge in an external electric field, what we most often calculate is not the electric potential energy but rather the electric potential. Electric potential We have seen that “electric potential energy” can include all the charge in a system – for example, we calculated the total potential energy of a system of three point charges. Or it can mean the potential energy of only some of the charges in an external electric field – for example, we calculated the potential energy of a dipole in a constant electric field. In the case of one point charge in an external electric field, what we most often calculate is not the electric potential energy but rather the electric potential. The relation between the electric potential energy of a point charge vs. the electric potential at the position of the point charge is exactly analogous to the relation between the electric force and the electric field. Electric potential We have seen that “electric potential energy” can include all charge in a system – for example, weiscalculated thefield totalat E(r) the electric Fthe q(r) is the electric force on energy of a system point charges. Or it can point r. apotential point charge q located at r. of threethe mean the potential energy of only one point charge in an external electric field – for example, we calculated the potential energy of each charge in a dipole inV(r) a constant electric potential field. is the electric U(r) is the potential energy at the point r. of a point charge q at r. In the latter case – the case of a point charge in an external electric field – what we most often calculate is not the electric The relation between the electric potential energy of a point charge vs. the electric potential at the position of the point charge is exactly analogous to the relation between the electric force and the electric field. Electric potential Fq(r) is the electric force on a point charge q located at r. E(r) is the electric field at the point r. U(r) is the potential energy of a point charge q at r. V(r) is the electric potential at the point r. The units of electric potential must be energy/charge. Our unit is the volt; 1 V = J/C = joules/coulombs. Electric potential Fq(r) is the electric force on a point charge q located at r. E(r) is the electric field at the point r. U(r) is the potential energy of a point charge q at r. V(r) is the electric potential at the point r. The units of electric potential must be energy/charge. Our unit is the volt; 1 V = J/C = joules/coulombs. Conversely, eV (the electron volt) is a unit of energy because it is charge (e) × potential (V). Electric potential Example: An electron accelerates from rest through an electric potential of 2 V. What is the final speed of the electron? Electric potential Example: An electron accelerates from rest through an electric potential of 2 V. What is the final speed of the electron? Answer: The electron acquires 2 eV kinetic energy. We have 1 2 19 J 19 me v [2 eV] 1.6 10 3.2 10 J , 2 eV and since the mass of the electron is me = 9.1 × 10–31 kg, the speed is v 2 3.2 1019 J 9.110–31 kg 8.4 105 m/s . Electric potential •The electric potential V(r) at a point r, due to a system of n point charges qi located at ri, is n V (r) i 1 qi . 4 0 ri r •The electric potential V(r) at a point r, due to a continuous charge density ρ(r), is (r ' ) V (r ) d r' 4 0 r 'r 3 . Electric potential Example 1: What is the electric potential at a point (x,y,z) due to a dipole made of charges q at (0,0,d/2) and –q at (0,0, –d/2)? z d/2 (x,y,z) –d/2 Electric potential Example 1: What is the electric potential at a point (x,y,z) due to a dipole made of charges q at (0,0,d/2) and –q at (0,0, –d/2)? Answer: z d/2 d 2 where r x y ( z ) 2 2 r– r+ –d/2 1 1 V ( x, y, z ) , 4 0 r r q (x,y,z) 2 . Electric potential Example 2: A rod with a uniform charge density λ lies along the z-axis between (0,0,–L/2) and (0,0,L/2). Let’s calculate the electric potential V(y) at a point (0,y,0) on the y-axis. z L/2 y –L/2 Electric potential Example 2: A rod with a uniform charge density λ lies along the z-axis between (0,0,–L/2) and (0,0,L/2). Let’s calculate the electric potential V(y) at a point (0,y,0) on the y-axis. L/2 V ( y) z L/2 y –L/2 dz 2 2 4 z y L / 2 0 4 0 L/2 2 2 ln z z y L/2 L / 2 L2 / 4 y 2 ln 4 0 L / 2 L2 / 4 y 2 . Halliday, Resnick and Krane, 5th Edition, Chap. 28, Prob. 4: The electric field inside a nonconducting, uniform sphere of radius R has magnitude E = qr/4πε0R3, where q is the total charge on the sphere and r is the distance from its center. (a) Given V(0) = 0, compute V(r). (b) What is the difference in electric potential between the surface of the sphere and its center? (c) Given V(∞) = 0, show that V(r) = q(3R2 –r2)/8πε0R3 inside the sphere. Why does this answer differ from the answer in part (a)? Halliday, Resnick and Krane, 5th Edition, Chap. 28, Prob. 4: The electric field inside a nonconducting, uniform sphere of radius R has magnitude E = qr/4πε0R3, where q is the total charge on the sphere and r is the distance from its center. (a) Given V(0) = 0, compute V(r). Answer: Symmetry dictates that E is radial. So we have to integrate E(r) = qr/4πε0R3 with respect to dr and, if necessary, add a constant to make V(0) = 0. Thus V (r ) drE(r ) dr and C = 0 yields V(0) = 0. qr 4 0 R 3 qr 2 8 0 R 3 C , Halliday, Resnick and Krane, 5th Edition, Chap. 28, Prob. 4: The electric field inside a nonconducting, uniform sphere of radius R has magnitude E = qr/4πε0R3, where q is the total charge on the sphere and r is the distance from its center. (b) What is the difference in electric potential between the surface of the sphere and its center? Answer: The question asks what is V(R) – V(0), and the answer is – q/8πε0R. Halliday, Resnick and Krane, 5th Edition, Chap. 28, Prob. 4: The electric field inside a nonconducting, uniform sphere of radius R has magnitude E = qr/4πε0R3, where q is the total charge on the sphere and r is the distance from its center. (c) Given V(∞) = 0, show that V(r) = q(3R2 –r2)/8πε0R3 inside the sphere. Why does this answer differ from the answer in part (a)? Answer: Outside the sphere, the potential must be V(r) = q/4πε0r, because the sphere looks just like a point charge q, and then we also satisfy V(∞) = 0. This outside potential must match the inside potential V(r) = q(3R2 –r2)/8πε0R3 at r = R and, indeed, both expressions yield V(R) = q/4πε0R. So V(r) is the same potential as in (a); they differ only by a constant so that in (a) we have V(0) = 0 and here we have V(0) = 3q/8πε0R.