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## Electric Potential and Electric Energy Chapter 17

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**Electric Potential and Electric Energy**Chapter 17**Potential Energy**Let's go back to junior physics for a second :) What is gravitational potential energy? Energy that depends on an object's mass and its position relative to some point i.e. To calculate someone's potential energy relative to the surface of the Earth you'd need mass, g and height above the surface**Electric Potential Energy**The idea of electric potential energy is similar to that of gravitational potential energy Electric potential energy for a charge is calculated based on the magnitude of the charge and its position relative to some point**Gravitational vs Electric Potential Energy (p. 504)**Caption: (a) Two rocks are at the same height. The larger rock has more potential energy. (b) Two charges have the same electric potential. The 2Q charge has more potential energy.**Recall point charges in electric fields (ch 16)**Let's say you have an electric field of magnitude 4500 N/C pointing toward the right If you place a proton in that field, what is the magnitude and direction of the force acting on that proton? F=qE=7.2 x 10-16 N Right + E**The proton in the electric field**So since the force acting on the proton is toward the right, it will accelerate toward the right What will happen to the proton's kinetic energy and electric potential energy? Kinetic Energy will increase EPE will decrease (conservation of energy)**Two charged plates (capacitor)**Let's say we've got two charged plates that are separated by a small distance (this is a capacitor) The E-field points from left to right + + + + + + + + + + + + - - - - - - - - - - - -**Two charged plates (capacitor)**A proton between these two plates would move towards the negative plate (right) An electron between these two plates would move towards the positive plate (left) + + + + + + + + + + + + - - - - - - - - - - - - E-field between Two charged plates**Two charged plates (capacitor)p. 503**A proton has the highest potential energy when it's near the positive plate An electron has the highest potential energy when it's near the negative plate + + + + + + + + + + + + - - - - - - - - - - - - E-field between Two charged plates**Potential for pos/neg charges**By convention, the positive plate is at a higher potential than the negative plate Positively charged objects move from higher potential to lower potential (i.e. towards negative plate) Negatively charged objects move from lower potential to higher potential (i.e. towards positive plate)**Electric Potential (V)**Electric Potential, V, is the potential energy per unit charge Unit is Volts (1 V= 1J/1 C) If a point charge, q, has an electric potential energy at some point a, then the electric potential is V= PE/q**Electric potential and Potential Energy**The change in potential energy of a charge, q, when moved between two points a and b ΔPE = PEb-PEa=qVba**Sample Problem p. 505**An electron in a television set is accelerated from rest through a potential difference Vba=+5000 V What is the change in PE of the electron? What is the speed of the electron as a result of the acceleration? Repeat for a proton that accelerates through a potential difference of -5000 V**Change in PE of electron**ΔPE = Peb-PEa=qVba ΔPE = qVba=(-1.6 x 10-19 C)(5000 V) ΔPE = -8 x 10-16 J Potential Energy was lost!**What is the speed of the electron as a result of the**acceleration? Conservation of Energy! The amount of PE lost, must be equal to the amount of KE gained! KE= 8 x 10-16 J=0.5mv2 V=4.2 x 107 m/s**For the proton**ΔPE = qVba=(1.6 x 10-19 C)(-5000 V) ΔPE = -8 x 10-16 J (Same as electron) Velocity is less because speed is greater V=9.8 x 105 m/s**Potential Difference**Since potential energy is always measured relative to some other point, only differences in potential energy are measurable Potential Difference is also known as voltage**Potential Difference**In order to move a charge between two points a and b, the electric force must do work on the charge Vab=Va-Vb= -Wba/q The potential difference between two points a and b is equal to the negative of the work done by the electric force to move the charge from point b to point a, divided by the charge**Sample Problem p. 522 #2**How much work is needed to move a proton from a point with a potential of +100 V to a point where it is -50 V?**Break it down**We're moving the proton from +100 V to -50 V Therefore point A is +100 V, point B is -50 V We're looking for the work done by the field -Wba= qVab=q(Va-Vb) -Wba= (1.6 x 10-19 C)(100V -(-50V)) Wba= -2.4 x 10-17 J**Back to the parallel plates!**For two parallel plates, the relationship between electric field and electric potential is below E=Vba/d d is the distance between the plates + + + + + + + + + + + + - - - - - - - - - - - - E-field between Two charged plates**The electron volt**The electron volt is another unit for energy 1 ev= 1.6 x 10-19 J Problem: A proton has 2 MeV of kinetic energy, how fast is it moving? 2 x106 eV= 3.2 x 10-13 J= 0.5mv2 V= 1.96 x 107 m/s**Section 17-Equipotential Lines**Equipotential lines are used to represent electric potential Equipotential lines are always perpendicular to electric field lines**Equipotential Lines (p. 507)**Equipotential lines (green) are perpendicular to the electric field lines (red)**17-5 Electric Potential due to Point Charges (p. 509)**The electric potential at a distance r from a single point charge q is : V=kQ/r Potential is zero at infinity The potential near a positive charge is large and decreases toward zero at large distances**Electric potential p. 509**The potential near a negative charge is negative and increases toward zero at large distances**Bringing charges togetherEx 17-3 p. 509**What minimum work is required by an external force to bring a charge q = 3.00 microC from a great distance away to a point 0.500 m from a charge Q= 20.0 microC?**Analyze the problem**Basically, we're taking the charge q from a place of zero potential, to a place of nonzero potential Use our trusty equation:Vab=Va-Vb= -Wba/q**Figure out the work done**The charge is coming from infinity, so Va=0 What is Vb? Vb=KQ/r=(9x109 Nm2/C2)(20x10-6C)/0.500m Vb= 360,000 V Wba= -q(Va-Vb)=-(3.00x10-6C)(0-360000V) W= 1.08 J**Electric potential of multiple charges**Electric fields are vectors, but electric potential is a scalar! When determining the electric potential at a point you can just add the electric potential from each charge, just be sure to include the correct sign of the charge when calculating potential**Example**Calculate the electric field at a point midway between a -0.5 microC charge and a -0.8 microC charge that are separated by 0.50 m. For the -0.5 microC charge, E= 72000 N/C left For the -0.8 microC charge, E= 115,200 N/C right Therefore E is 43200 N/C right**Electric Potential**Calculate the electric potential at a point midway between a -0.5 microC charge and a -0.8 microC charge that are separated by 0.50 m. For the -0.5 microC charge, V=kQ/r= (9x109 Nm2/C2)(-0.5 x 10-6 C)/0.25m V= -18000 N/C**Electric Potential**For the -0.8 microC charge, V=kQ/r= (9x109 Nm2/C2)(-0.8 x 10-6 C)/0.25m V= -28800 V Total V= -46800 V This is much easier! No directions...just make sure you include the sign!**+**+ + + + + + + + + + + - - - - - - - - - - - - E-field between Two charged plates Section 17-7- Capacitance • A capacitor stores electric charge and consists of two conducting objects that are placed next to each other but not touching**+**+ + + + + + + + + + + - - - - - - - - - - - - Capacitance p. 513 • If a voltage is applied to a capacitor (i.e. connected to a battery), then it becomes charged • Amount of charge for each plate: • C= Capacitance of the capacitor (different for each capacitor) • Unit for C is farad (F)**+**+ + + + + + + + + + + - - - - - - - - - - - - d Capacitance of the Capacitor • A= Area of plates • If A increases, C increases • d= distance between the plates • If d increases, C decreases • ε0 = 8.85 x 10-12 C2/Nm2 • (This is the permitivity of free space)**Storage of Electrical Energy**• A charged capacitor stores electric energy**Sample Problem p. 524 #41**• A 7.7 µF capacitor is charged by a 125 V battery and then is disconnected from the battery. When this capacitor (C1) is connected to a second, uncharged capacitor (C2), the voltage on the first drops to 15 V. What is the value of C2? (Charge is conserved)**Solve the Problem**• For the first capacitor: • When the capacitors are connected, the voltage on the first one is 15 V. That means the new charge on C1 is:**Solving the problem**• What happens to the rest of the charge? • It must be on capacitor 2 because charge is conserved • Since the two capacitors are connected, the voltage for the second one must also be 15 V**Connected Capacitors**• Capacitors can be connected in series or parallel • When capacitors are connected in parallel, the equivalent capacitance is the sum • The voltage across each capacitor is the same**Capacitors in Series**• If the capacitors are connected in series, the equivalent capacitance is given by the following expression**Capacitors in Series**• For capacitors in series, the total voltage must equal the sum of the voltages across each capacitor • The charge on each capacitor is the same as the charge on the equivalent capacitor for capacitors in series**Sample Problem**• What is the equivalent capacitance for this combination of capacitors? • C2 and C3 are connected in parallel • Combine them into one capacitor • C23=C2 + C3 = 35 µF C1 = 12 µF C2 = 25 µF C3 = 10 µF**Simplify the Combination**• C23 and C1 are connected in series C1 = 12 µF C23 = 35 µF**Sample Problem Continued**• How much charge is stored on each capacitor? • Q=CV • V1= 50 V (this is the voltage across C1)**Sample Problem Continued**• C1 and C23 are connected in series, therefore the charge on C23 is the same as the charge on C1**Sample Problem Continued**• C2 and C3 are connected in parallel, therefore: