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Review of Electrical Potential (A makeup for yesterday) Note on equations General equation (always works as long as you do the calculus right) V V E d b b a a Specific equation for a point charge with infinity as kq your reference point (i.e. where V = 0) V= Specific r equation for finding the potential difference in a uniform, constant electric field (i.e. the field from an infinite plane of charge) V Ed (d is the distance between your reference point and the point you are interested in) Ball on a hill analogy for potential energy U U = mgh h Remember U =0 that you can choose your zero of potential (U) to be anywhere Here we choose U = 0 at the bottom of the hill So ∆U = Ufinal – Uinitial = -mgh And the change in kinetic energy ∆ KE = -∆U = mgh (i.e. the ball has a real speed at the bottom of the hill as the potential energy is converted into kinetic energy) Ball on a hill analogy for potential energy U U=0 h Now U = -mgh we choose U = 0 at the top of the hill So ∆U = Ufinal – Uinitial = -mgh And again the change in kinetic energy ∆KE = -∆U = mgh Ball on a hill analogy for potential energy U The important point is that nature always wants to be in the lowest potential energy configuration ◦In this case that means the ball wants to roll down hill (amazing how much physics can complicate the obvious, eh?) Relating the ball on a hill to E&M Consider a positive charge +Q fixed at the origin with a small test charge +q close by Intuitively, we +q +Q know +q will be repelled from +Q, just like a ball is repelled from the top of a hill In physics-speak, +q is forced towards a lower potential energy (U) or the charge wants ∆U to be negative Finally, to relate this to the electrical potential (V), U = qV, so a positive test charge +q will move towards lower V (in other words it wants ∆V to be negative) Relating the ball on a hill to E&M Now consider the same thing but with a negative test charge -q Intuitively, we -q +Q know -q will be attracted to +Q In physics-speak, -q is forced towards a lower potential energy (U) or the charge wants ∆U to be negative Finally, to relate this to the electrical potential (V), U = qV, so a negative test charge -q will move towards higher V (in other words it wants ∆V to be positive) One last clarification Going back to the first case with two positive charges r +q +Q To flesh this out a little more, if we wanted to find the change in electrical potential (a.k.a. if we wanted to find the potential difference ∆V), we first find the potential from +Q because this creates the field through which we move the test charge +q V kQ r One last clarification r +q +Q So the potential difference is then And kQ kQ V rfinal rinitial the change in potential energy is kQ kQ U q r r final initial So if the charge is repelled, rfinal > rinitial and ∆U is then negative like we expected from the ball on a hill analogy Think about this on your own for the case of a negative test charge (you should still get a negative change in potential energy for the –q being attracted to +Q)