Today 3/12  Plates if charge  E-Field  Potential  HW:“Plate Potential” Due Friday, 3/14

 (+) How big is E?  (-) E net = 0

 0 = Q/A Charged conducting plate What’s wrong with this picture? A = Area of one side

 L =  0 /2 Charged conducting plate Free charge always goes to surface of conductor.  R =  0 /2

 L =  0 /2 Charged conducting plate  R =  0 /2 What is the electric field inside the conductor? E L =  L /2  0 =  0 /4  0

 L =  0 /2 Charged conducting plate  R =  0 /2 What is the electric field inside the conductor? E R =  R /2  0 =  0 /4  0

 L =  0 /2 Charged conducting plate  R =  0 /2 What is the electric field inside the conductor? The electric field is zero everywhere inside the conductor. Always, any conductor, no exceptions.

q=+1C What happens if I let it go?

KE = 0KE = 100 J A B Assume the particle gains 100 joules of kinetic energy as it moves from A to B. q=+1C

A B Now I stop it at B. How much work must I do to move it back to A? +100 J How does the potential energy change in moving from B to A? +100 J  PE BA = +100 J  PE AB = -100 J q=+1C

A B  PE BA = +200 J  PE AB = -200 J What if q=+2C? How much work must I do to move it back to A? How does the potential energy change in moving from B to A? +200 J q=+2C

A B Now we are back to our original definition.  PE BA = (+100 J/C)x(q)  V BA tells us how much PE changes when +1C is moved from B to A.  PE BA =  V BA q  V BA = +100 J/C q=+1C

A B What if q = -1C? First I must turn my hand around. q= -1C

A B What if q= -1C? How much work must I do to move it back to A? -100 J How does the potential energy change in moving from B to A? -100 J  PE BA = (+100 J/C)x(q)  PE BA = (+100 J/C)x(-1)  V BA tells us how much PE changes when +1C is moved from B to A.  PE BA =  V BA q q= -1C

A B What if q= -1C? How much work must I do to move it back to A? -100 J How does the potential energy change in moving from B to A? -100 J  PE BA = (+100 J/C)x(q)  PE BA = (+100 J/C)x(-1)  V BA does not depend on the sign of the point charge but  PE BA does!!!!!  PE BA =  V BA q q= -1C

A B  V AB = -100 J/C   V AB = -100 volts A proton is released from rest at A. What is its speed when it reaches B? m= 1.7 x kg q= 1.6 x C  PE AB = q  V AB  PE AB = q  -100 J/C)  PE AB = -1.6 x J What happens to the kinetic energy?

A B  KE AB = +1.6 x J  A proton is released from rest at A. What is its speed when it reaches B? m= 1.7 x kg q= 1.6 x C  mv 2 = +1.6 x J v = 1.4 x 10 5 m/s What happens to the kinetic energy?  PE AB = -1.6 x J

A B A B  What direction is the force on an electron? EF

A B  V AB = -100 J/C   V BA = +100 J/C An electron is released from rest at B. What is its speed when it reaches A? m= 9.1 x kg q= -1.6 x C  PE BA = q  V BA  PE BA = q  +100 J/C)  PE BA = -1.6 x J What happens to the kinetic energy?

A B  KE BA = +1.6 x J   mv 2 = +1.6 x J v = 5.9 x 10 6 m/s An electron is released from rest at B. What is its speed when it reaches A? m= 9.1 x kg q= -1.6 x C What happens to the kinetic energy?  PE BA = -1.6 x J

B A   V AB = ?

B A  How does doubling the E-field affect  V AB ?  V AB = ?  V AB doubles

B new A  How does moving point B affect  V AB ?  V AB = ? How does  V AB depend on E and D? D Anything else?  V AB is halved  B old

B A  KE AB = 0 constant speed v0v0 v0v0 KE A =1/2 mv 0 2 KE B =1/2 mv 0 2 How much work must I do to move the charge from A to B? D

B A F E = qEF Hq W AB = F E x D W AB = qED W AB = F Hq x D

B A What is the change in potential energy in going from A to B? F E = qE F Hq W AB = qED PE AB = qED PE AB = q  V AB  V AB = ED AB W AB = qED Only applies when the field is uniform over the distance.  V AB ‘s sign depends on the direction of E. In this case it’s positive.