Chapter 25 Review: Electric Potential Matthew Volk Background / Summary Electric potential is the potential energy per unit charge of a charged object in an electric field. This chapter covers potential energy, electric potential difference (voltage), voltages in constant electric fields, voltages due to point charges, voltages due to continuous charges, and the relationship between voltage and electric field strength. For a quick summary, look at the formulae table, but for a more in-depth description of each equation look at the topic summaries. Potential Energy and Voltage The change in potential energy of a charged object moving through an electric field is given by the equation: Formulae to Know: Voltage is defined as the change in potential energy per unit charge: Work is given by the following equation: Voltages in Constant Electric Fields For constant electric fields, E can be removed from the integral: Protons in Electric Fields High V High U Low V Low U Field lines – Field lines always go from high to low voltage. Positive charges follow them, going from high U and high V to low U and low V. Negative charges move in the opposite direction as them, going from high U to low U, but low V to high V (remember is negative). You can use these equations to find velocity of a particle in an electric field using conservation of energy: Electrons in Electric Fields High V Low U Low V High U Voltage due to Point Charges Voltage due to a point charge is given by the following relationship. For multiple point charges, you sum up the resulting from each charge. ’s Millikan Oil Drop Experiment qE mg Potential energy of a system of particles is given by similar equations. For two particles, the first equation applies. For 3 or more, the equation needs to be expanded as shown: Definitions to Know: Voltage (Electric Potential) – defined as the potential energy per unit charge of an object in an electric field Electric Field Lines – Always point in the direction of decreasing voltage Equipotential Surfaces – surfaces over which all points have the same electric potential Millikan Oil Drop Experiment – Experiment that proved charge is quantized and calculated the elemental charge unit Tip: For electric potential due to various common charge distributions, like spheres or charged rings, look at table 25.1 (page 786) in the book. It gives a comprehensive list of the results of the equations given here that might be useful to memorize if you prefer the memorization approach. I recommend memorizing the sphere ones at the least, because it’s useful to not have to derive them each time. Electric Field and Voltage The magnitude of the electric field is equal to the opposite of the change in voltage over a set distance. In other words, For objects in 2D or 3D space, you have to take partial derivatives, or you can use the Del operator for shorthand: Voltage due to a Continuous Charge Distribution To find the voltage change due to a continuous charge distribution, you take the equation used for multiple charges and change the sigma to an integrand: Miscellaneous Stuff to Know The voltage is constant across the surface of a charged conductor. In other words, The Millikan Oil Drop Experiment – This experiment is famous because it proved that charge is quantized and measured the charge of an electron. Essentially, they sprayed ionized particles of oil into a variable electric field controlled by a dial. They would change the field strength until some particles remained suspended or stopped accelerating, in which case they knew that the force of gravity was equal to the force of the electric field: From there, they calculated q on many different particles, found the smallest q to equal qe, and plotted them to find a stepwise function, proving that charge is quantized. Problems 1. [Easy] Three point charges are located along the x-axis as follows: a +2.00 μC point charge is located at + 3.00 mm, a 0.50 μC point charge is located at – 5.00 mm, and a +5.00 μC point charge is located at the origin. a) What is the potential energy of this system of point charges? b) What is the resultant voltage at x=1.00 cm due to this system of point charges? 2. [Medium] A particle enters an electric field with an initial velocity to the left of 1.50 x 10 6 m/s. The electric field lines point to the right. Assume there is no loss of energy. What voltage drop must this particle travel through before its velocity reaches 0 if it is a proton? If it is an electron? C B [Hard] Two spherical conducting shells are arranged as shown in the figure to the right. The inner radius 3. r1 = 5.00 cm and has charge 4.00 μC. The outer shell has radius r2 = 10.0 cm and has charge -6.00 μC. a) What is the electric field in regions A, B, and C? b) What is the electric potential V in regions A, B, and C? A r1 r2 Problem 1 q1 = +0.50 μC q2 = +5.00 μC q3 = +2.00 μC a potential energy, I use the equation for potential energy of a system of particles: a) To find b) To find voltage, I use the voltage equation: Problem 2 a) This is an energy problem. So, start with the conservation of energy equation for electric field problems: b) Because an electron will accelerate in the direction opposite that of the field lines, it will speed up through this field, not slow down. Therefore, no change in voltage will cause this electron to stop in this field. (Not possible.) Problem 3 a) In area A, we know from Gauss’s Law that, because there is no charge internal, there is no electric field. So, C B In area B, we can apply the equation for the electric field outside a sphere (see ch. 24 if you need help with Gauss’s Law) A r1 r2 In area C, we use the same equation, noting that the total enclosed charge is now the sum of the two charges. b) In area C, we can use the following equation to get a function of V(r): VB equals VC at r2 plus the effects of the voltage in area B as well. In other words, we simply add the voltage at r2 to the integral of the electric field equation derived in part a: Finding VA is significantly easier. The voltage inside area A is the same as the voltage at points a distance r1 from the center. Because r1 is within the domain of the equation we just created for VB, we can find VA by taking . This makes sense because we know that the voltage inside area A should be constant.