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 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)

Potential for positive/negative charges

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 Δ PE = PE b -PE a =qV ba

Sample Problem p. 505 An electron in a television set is accelerated from rest through a potential difference V ba =+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 V

Change in PE of electron Δ PE = Pe b -PE a =qV ba Δ PE = qV ba =(-1.6 x C)(5000 V) Δ PE = -8 x 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 J=0.5mv 2 V=4.2 x 10 7 m/s

For the proton Δ PE = qV ba =(1.6 x C)(-5000 V) Δ PE = -8 x J (Same as electron) Velocity is less because speed is greater V=9.8 x 10 5 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 V ab =V a -V b = -W ba /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 -W ba = qV ab =q(V a -V b ) -W ba = (1.6 x C)(100V -(-50V)) W ba = -2.4 x J

Back to the parallel plates! For two parallel plates, the relationship between electric field and electric potential is below E=V ba /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 J Problem: A proton has 2 MeV of kinetic energy, how fast is it moving? 2 x10 6 eV= 3.2 x J= 0.5mv 2 V= 1.96 x 10 7 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 together Ex 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 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:V ab =V a -V b = -W ba /q

Figure out the work done The charge is coming from infinity, so V a =0 What is V b ?  V b =KQ/r=(9x10 9 Nm 2 /C 2 )(20x C)/0.500m  V b = 360,000 V W ba = -q(V a -V b )=-(3.00x10 -6 C)( V) 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= N/C left For the -0.8 microC charge, E= 115,200 N/C right Therefore E is 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= (9x10 9 Nm 2 /C 2 )(-0.5 x C)/0.25m V= N/C

Electric Potential For the -0.8 microC charge, V=kQ/r= (9x10 9 Nm 2 /C 2 )(-0.8 x C)/0.25m V= V Total V= V  This is much easier! No directions...just make sure you include the sign!

Section Capacitance A capacitor stores electric charge and consists of two conducting objects that are placed next to each other but not touching E-field between Two charged plates

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)

Capacitance of the Capacitor A= Area of plates If A increases, C increases d= distance between the plates If d increases, C decreases ε 0 = 8.85 x C 2 /Nm 2 (This is the permitivity of free space) d

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 (C 1 ) is connected to a second, uncharged capacitor (C 2 ), the voltage on the first drops to 15 V. What is the value of C 2 ? (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 Parallel

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 C 23 =C 2 + C 3 = 35 µF C1 = 12 µF C2 = 25 µF C3 = 10 µF

Simplify the Combination C 23 and C 1 are connected in series C 1 = 12 µF C 23 = 35 µF

Sample Problem Continued How much charge is stored on each capacitor? Q=CV V 1 = 50 V (this is the voltage across C1)

Sample Problem Continued C 1 and C 23 are connected in series, therefore the charge on C 23 is the same as the charge on C 1

Sample Problem Continued C2 and C3 are connected in parallel, therefore:

Sample Problem Finished! The charge on C2 is: The charge on C3 is: