Electric Potential Physics 102 Professor Lee Carkner Lecture 11
PAL #10 Find E 1 and E 2 E 1 = 8.99X10 9 (2X10 -6 )/(3 2 ) = 1124 N /C r 2 2 = r 2 = 5 m E 2 = 8.99X10 9 (4X10 -6 )/(5 2 ) = 1438 N /C Find 1 and 2 1 = 0 (right on x-axis) Can get 2 from triangle tan 2 = ¾, 2 = 37 degrees 2 is below X axis so is m 3 m q 2 = +4 C q 1 = +2 C r 2 =5 m E2E2 E1E1 22 22
PAL #10 Find X and Y components E 1x = E 1 cos 1 = 1124 cos 0 = 1124 N/C E 1y = E 1 sin 1 = 1124 sin 0 = 0 E 2x = E 2 cos 2 = 1438 cos -37 = 1148 N/C E 2y = E 2 sin 2 = 1438 cos -37 = N/C Find total E vector and angle E total,x = E 1x + E 2x = = 2272 N/C E total,y = E 1y + E 2y = 0 – 865 = -865 N/C E 2 total = E 2 total,x + E 2 total,y E = ( ) ½ E = 2431 N/C tan total = E total,x / E total,y = -865 / 2272 total = -20 degrees 20 degrees below + X axis 4 m 3 m q 2 = +4 C q 1 = +2 C r 2 =5 m E2E2 E1E1 22 22
Electrical Potential Energy The value of the potential tells us how much potential energy is at that point per unit charge V = PE/q or PE = Vq 1V = 1 J/C Which could be converted into work or kinetic energy
Point Charges and Potential Consider a point charge Q, what is the potential for the area around it? At infinity the potential is zero It can be shown that: V = k e Q / r Q is charge providing potential, q is charge experiencing potential
Finding Potential If the potential is provided by an arrangement of charges: Total V = V 1 + V 2 + V 3 … V = -Ed We don’t know the potential at anyone point, but we can find the difference between two points separated by a distance d
Potential and Energy The change in energy is: PE = V f q – V i q = q(V f -V i ) = q V PE = -W Energy could be converted into kinetic energy: PE = KE KE i + PE i = KE f + PE f Particle could speed up or slow down
Problem Solving Particle moving between two points of different potential Find V and then PE (=q V) Add or subtract PE to initial energy of particle
Signs As a positive charge moves with the electric field, the particle gains kinetic energy and loses potential energy Like dropping a ball in a gravitational field the electric field does positive work If a positive charge is forced backwards against an electric field, the particle loses kinetic energy and gains potential energy like rolling a ball up a hill the field “does” negative work
Potential and Charges A negative particle: Opposite of positive particle A negative particle has the most potential energy at low potential The potential and the potential energy are two different things Potential at a point is the same no mater what kind of test charge is put there
High potential Low potential
Work Since energy must be conserved: PE + W = 0 or PE = -W Work done by the field is positive if it decreases the potential energy The “natural” movement When the charge is forced in the “unnatural” direction The negative work done by the system is the positive work done on the system
E + Down gain KE lose PE field does +work “natural” Up lose KE gain PE you do work field “does” negative work “forced” For negative particle, everything is backwards But high and low potential are still in the same place High potential Low potential
Equipotentials Each line represents one value of V Particles moving along an equipotential do not gain or lose energy Equipotentials cannot cross Blue = field = E Dashed = potential = V
Next Time Read Ch Homework, Ch 17: P 10, 20, 35, 46
Electric Potential Chart sign of PE sign of V sign of Wnaturally? + charge moves with E field + charge moves against E field -charge moves with E field -charge moves against E field
The above electric field, A)increases to the right B)increases to the left C)increases up D)increases down E)is uniform
Is it possible to have a zero electric field on a line connecting two positive charges? A)Yes, at one point on the line B)Yes, along the entire line C)No, the electric field must always be greater than zero D)No, but it would be possible for two negative charges E)No, the electric field is only zero at large distances