Week 5 Class Monday (Finish chapter 29), Wednesday, Friday CLASS M,W,F Electric Potential Chapter 29 In intro classes I try to develop an intuitive approach to Voltage by examining the way gravitational PE/unit mass is described. • PE depends on h [height] • All height information is based on Δh=hfinal-­‐hintitial. Really concerned with Voltage differences. • Implies an arbitrary point that can be declared = 0 è GND • Height causes the flow of water (rivers), bikers, skiers… • No matter how you get to a particular height the PE is the same o Parachute from a plane and land on the physics roof, get launced via a catapult or take the stairs. Every one on the roof has the same PE. o Gravity and Coulomb force are conservative. o
!"#$%& !""#
𝐹 ∙ 𝑑𝐴 = 0 Potential is a scalar NOT a vector. So the total PE at a point is the sum of the PE of all the charges. 1 𝑄!
𝑃𝐸 =
𝑃𝐸! =
𝑞 !
! 4𝜋𝜖! 𝑟!
1 𝑄!
𝑉=
𝑉! =
!
! 4𝜋𝜖! 𝑟!
PE of a dipole in a constant electric field 𝑈 = −𝑝 ∙ 𝐸 PE ring of chargeè Voltage when per unit charge V of a disk V of a point charge V of parallel plate Conductorsè If you argue that the charges inside a conductor are free to move but not leave the conductor, then you must build up a surface charge that cancels the electric field inside. If there was a field then the charges in the field would move so a stable equilibrium requires Vinside=0. PE of a dipole The PE energy derived for a dipoles addresses the specific and limited aspect of the rotation of the dipole about a fixed point where the field is 𝐸 . •
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Only allow the dipole to rotate. {Special case where the dipole is not allowed to translate or move to a new point in space.} Assume that the dipole is small so that the forces that produce the Torque are just 𝐹 = 𝐸𝑞 where 𝐸 is the field at the dipole. Define Θ as the angle between the dipole and the field. Calculate the torque as 𝜏 = 𝑟𝑥𝐹 = 𝑝𝐸𝑠𝑖𝑛𝜃 [𝑝𝑒𝑟𝑝. 𝑡𝑜 𝑡ℎ𝑒 𝑝 − 𝐸 𝑝𝑙𝑎𝑛𝑒] Choose one charge to be the point about which the torque is calculated or choose the dipole center. Find the projection of the separation 𝑠 𝑠𝑖𝑛𝜃. Use the work formula for rotating via a torque 𝑊 = 𝜏𝑑𝜃 = [𝜏 ∙ 𝑑𝜃] must rotate about the axis defined by the torque if you remove the vector aspect. This it the appropriate rotation for the dipole. Remember that to calculate potential energy you need to find the torque that is applied to rotate the dipole against the field torque (minus sign). Upcoming topics: Chapter 30 potential and field 𝑉=−
𝐸 ∙ 𝑑𝑠 !
parallel plate 𝐸 = ! !
!"
𝐸 = − !" (steepest uphill) Made a connection between contour lines on a map and gravitational PE and PE/m. 𝑈!"#$%&' = 𝑚𝑔ℎ 𝑈!"#$%&'
= 𝑔ℎ = 𝑉!"#$%&' 𝑚
Walking along a contour you are not going up or down. Crossing many contour lines in a short distance (viewed from top) shows that there is a steep rise or decent while widely spaced line show gentle slopes. You can proceed on a hill to climb following the steepest direction or use switchbacks that do not follow the steepest direction. At any point on a hill a person can find the direction of no change, the steepest uphill and the steepest downhill. !"
These directions are significant because !" is just a measure of this. !"
!"
!"
!"
!"
!"
= 0 along a contour = + climing uphill = -­‐ going down hill Remember that differentials are simple small deltas. 𝑑𝑉 Δ𝑉
Δ𝑉
=
𝑙𝑖𝑚𝑖𝑡 𝑡ℎ𝑎𝑡 Δ𝑠 𝑔𝑜𝑒𝑠 𝑡𝑜 0 = lim
!!→! Δ𝑠
𝑑𝑠 Δ𝑠
So you can imagine taking a small step and the seeing how Δℎ changes. Gravity always point down. But the combination of a surface and gravity generates an effective field. The normal force + gravity force is the relevant force for this model. Thinking about hills and the relationship between steepness helps to understand the relationship between fields and potentials. If we want to be perfectly clear we need to understand that steepness on a hill is an effective way of thinking about contours and changing contours but gravity by itself is doesn’t change the slope of the hill changes and that adds more or less gravitational force because the normal of the hill is less effective. Emphasize again that all potentials are changes. There is no absolute reference but often the obvious reference point is chosen so it is not always identified specifically but there is always a chosen point where U=0 or V=0. Conductors E must be normal to the surface Sharp points have large fields. Capacitors 𝑄 = 𝐶𝑉 𝑄
𝑉=
𝑑 𝜖! 𝐴
Gauss’s Law can be used to derive the field. {Very quick example give}. Friday Emphasize the difference between V for a capacitor and Energy stored in a capacitor. V is the energy per unit charge that it takes to move from one side to the other when the capacitor is charged. E is the energy required to charge up the capacitor. Back to hills è It takes a certain amount of energy to pile up the dirt to build a mountain (charge up the capacitor). You gain energy as you ski down the mountain (drop across a capacitor). The two quantities are not the same. Moving the charge onto the capacitor is a process that starts with Q=0. So the first charge is free. Placing the next charge is more difficult and placing the last charge is the most difficult. Reviewed the rules for a conductor. E=0 inside E= perpendicular to the surface Charge moves to cancel internal field Charge is free to move. Charge may not leave the conductor. Chapter 31 Resistance, Capacitance with DC power supplies, current Chapter 32 More basic circuits Chapter 33 Magnetic fields, materials, forces and sources Chapter 34 Induction Chapter 35 waved and fields è LIGHT, Maxwell’s equations Chapter 36 AC circuits