Welcome Back to Physics Electric Potential. Robert Millikan Physics 1308: General Physics II - Professor Jodi Cooley

Transcription

1 Welcome Back to Physics 1308 Electric Potential Robert Millikan

2 Announcements Assignments for Tuesday, September 18th: - Reading: Chapter Watch Video: Lecture 8 - Capacitors and Simple Circuits Homework 4 Assigned - due before class on Tuesday, September 18th. Midterm Exam 1 will be in class on Thursday, September 20th. Jasmine s office hours have been changed. To accommodate students better, they will be MONDAYS ONLY from 2-3:50 pm. Location: FOSC 153 (same place).

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4 Review Question 1 A positively-charged particle is held at point A between two parallel metal plates. The plate on the left has a net positive charge +q and the plate on the right has a net negative charge q. The particle is then moved to point B. How does the electric potential energy at point A compare with that at point B? A) UA > UB B) UA = UB C) UA < UB D) UA > UB or UA < UB depending on the actual distances from the points to the plates.

5 Key Concepts The electric potential due to a group of charges: is the algebraic sum of the individual potentials with no consideration of direction. The electric potential energy of a system of charged particles: is equal to the work needed to assemble the system with the particles initially at rest and infinitely distant from each other.

6 Key Concepts Electric charges exert forces on one another via the electric field, which is always present whether or not a charge is there to by acted upon. The electric force is a conservative force field with an associated electric potential energy. Energy can be stored in and released from an electric field. The energy available to do work per unit charge is the Electric Potential - it exists whether or not there is a charge present in the electric field or not.

7 Instructor Problem: What s the Difference? B Calculate the difference in electric potential between points A and B for the point-charge electric field (due to a single electron) shown to the left. Point A is a distance of 1.0 x 10-8 m from the electron while Point B is a distance of 4.0 x 10-8 m from the electron. A

8 First sketch the problem and take account of what is known. Given: Find: Single electron! q = 1 e = C V BA r A = m r B = m We could find the work to move from a charge from A to B directly, but that is complicated. Instead, consider that by definition, V BA = V B V A =(V B V 1 ) (V A V 1 ) =0 =0 Note: We have added a pair of zeros, but no constant this helps because this helps us to see that this one potential difference is just really only have a pair of potential differences.

9 We know that V A V 1 = k 1 r A V B V 1 = k 1 r B Thus, V BA = k q r B k q r A = kq( 1 r B 1 r A ) =( N m 2 /C 2 )( C)( m m ) =0.11 N m C =0.11 J C =0.11 V! V BA =0.11 V

10 Student Problem: Ions Near a Membrane - Revisited How much energy is stored in the arrangement of a potassium ion, a sodium ion, and the anion shown to the left? Answer the question using electron-volts (ev) as the energy unit. HINT: Assume the ions were brought together from an infinite distance away, and treat them as point charges.

11 First sketch the problem and take account of what is known. Given: Find: q A = 4e Utotal q Na =1e q K =1e L = m We need to find the energy of assembly the energy to bring all the charges together. This is the energy requited to bring all the charges from infinity. U total = q A V A + q Na V Na + q K V K = U = U f U 1 =0 J So, lets just bring in the charges from infinity, one at a time, building the system.

12 1) A It cost no energy to bring A from r = infinity if there are no other charges present. So, VA = 0. 2) Na Now bring in sodium. Here we must calculate the potential energy of the sodium- anion pair. U Na = q Na V Na = q Na k q A r A,Na = k q Naq A 2L 3) K Finally, we bring in the potassium. To do so, we must do work to bring the potassium near both the anion and the sodium, thus account for both the potential of the anion and the sodium. U K = q k (V K,A + V K,Na ) = q k k q A + k q Na r A,K r Na,K r A,K =2L r Na,K = p 8L ` = p (2L) 2 +(2L) 2 = k q Kq A 2L + k q Kq Na p 8L

13 Now put it all together: U = k q Naq A 2L + k q Kq A 2L + k q Kq Na p 8L plug in numbers. U = J 1eV = J U = 1.9 ev

14 The End for Today!