Electric Potential

Gravitational Potential Energy B hBhB F = mg hAhA A GPE = GPE = required to raise or lower the book. - Where =

Electric Potential Energy F = q o E A dAdA + B dBdB ΔEPE = -W E(AB) = Does a proton at rest at point A have more or less potential energy than it would at point B? More

Electric Potential Energy of Point Charges Much like the book is attracted to the earth due to gravity, two unlike charges are attracted to one another. Conversely, like charges repel. It takes to move away from one another and to move them closer together. F = kqq o r2r2 U e = Fr = +q -q o r E

Electric Potential Energy and Work of Point Charges +q -q o rArA A +q -q o rBrB B To change the energy level from U A to U B, it requires work (W). -W = –.

Electric Potential Energy 1.What would happen if the charged particle q was fixed in place and then particle q o was suddenly released from rest? A. It would accelerate away from q. B. It would accelerate towards q. C. It would stay where it is. 2.How would the potential energy of this system change? A. It would increase. B. It would decrease. C. It would remain the same. +q -q o

Electric Potential SI Units: / = 1 ( ) The Electric Potential Difference is equal to the required to move a test charge from infinity to a point in an divided by the of the test charge. The Electric Potential is the per unit of ( / ).

Relationship Between Electric Potential and Distance(point charges) Consider relationship between V and r. What happens to V as r B goes to ? As r increases, i.e., as r B  , V . The relationship above reduces to: V = / The of the charge will determine if the electric potential is or. When two or more charges are present, the total electric potential is the from all the charges present in the system. V B - V A = = -

Electric Potential(point charges) Consider the following system of three point charges. What is the electric potential that these charges give rise to at some arbitrary point P? Use superposition to determine V. Note that the electric potential can be determined from any point in space. P Q1Q1 Q3Q3 Q2Q2 r1r1 r2r2 r3r3 + +V =

Electric Potential and Electrical Potential Energy/Work (point charges) If we now move a test charge from infinity to point P, we can determine the potential energy of the system or the work required to the test charge to its new location. Remember: =. qoqo Q1Q1 Q3Q3 Q2Q2 r1r1 r2r2 r3r3

Example 1: Two Point Charges Two point charges, µC and µC, are separated by 1.00 m. What is the electric potential midway between them? A μC B μC 0.5 m

Characteristics of a Capacitor Uniform Electric Field Two equal and oppositely charged plates E Since the is, the force acting on a charged particle will be the everywhere between the plates. F e = B qoqo qoqo A qoqo C

Electric Potential and Work in a Capacitor qoqo A F = q o E qoqo B dAdA dBdB W AB = - W AB =  V = = If W AB = q o Ed, then what is W CD ? qoqo D C W CD = Joules because the acts to the of. Do you remember that W = ?

Electric Potential of a Capacitor – An alternative From mechanics, W =. From the previous slide, W =. From the reference table, V =. V = Uniform Electric Field qoqo A F = q o E B d Two equal and oppositely charged plates

Example 2:Parallel Plates A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.50 mm. When an electric spark jumps between them, the magnitude of the electric field is 4.8 x 10 7 V/m. What is the magnitude of the potential difference V between the conductors? d

Example 3: Parallel Plates A proton and an electron are released from rest from a similarly charged plate of a capacitor. The electric potential is 100,000 V and the distance between the two plates is 0.10 mm. 1. Which charge will have greater kinetic energy at the moment it reaches the opposite plate? 2. Determine the amount of work done on each particle. 3. Determine the speed of each particle at the moment it reaches the opposite plate. 4. Determine the magnitude of the force acting on each particle. 5. Determine the magnitude of the acceleration of each particle.

Example 3: Parallel Plates(cont.) Begin by drawing a picture and listing what is known: V = d = q e = p+p+ e-e- d

Example 3: Parallel Plates(#1 & #2) For #1, you could answer #2 first to verify. The answer is that the of particles will be the Why? because of the formula needed in question #2 applies to both charges, and =. Hence:

Example 3: Parallel Plates(#3) Apply the to determine the final speed of the electron and proton. Since the is equal to : Proton: Electron:

Example 3: Parallel Plates(#4) Since F =, it will be the for both particles because their are the and the electric field is between two parallel plates. We also know that W =. Since we know the between the and the done to move either charge from one plate to another, we can determine the force as follows:

Example 3: Parallel Plates(#5) Since we have the acting on each particle, we can now calculate the of each particle using.

Equipotential Lines Equipotential lines denote where the electric potential is the in an electric field. The potential is the anywhere on an a distance from a point charge, or from a plate. No is done to move a charge an. Hence = (The electric potential difference does depend on the taken from to ). Electric field lines and equipotential lines cross at and point in the direction of potential.

Equipotential Lines Parallel Plate Capacitor Electric Potential / Voltage Note:

Equipotential Lines Point Charge + Electric Potential / Voltage Note: Note: A charged surface is also an !

Equipotential Lines (Examples) EField/EField.html EField/EField.html

Key Ideas Electric potential energy (U) is the work required to bring a positive unit charge from infinity to a point in an electric field. Electric potential (V) is the change in energy per unit charge as the charge is brought from one point to another. The electric field between two charged plates is constant meaning that the force is constant between them as well. The electric potential between two points is not dependent on the path taken to get there. Electric field lines and lines of equipotential intersect at right angles.

Electric Potential Energy and Work in a Uniform Electric Field qoqo A F = q o E qoqo B dAdA dBdB W AB = EPE B – EPE A W AB = Fd B – Fd A W AB = q o Ed B – q o Ed A W AB = q o E(d B – d A ) = q o Ed Note: The force acting on the charge is constant as it moves from one plate to another because the electric field is uniform.