Lesson 58 - Potential Difference and Electric Potential Part 2

1. Lesson 58 - Potential Difference and Electric Potential Part 2 By: Fernando Morales, Brian Cox’s Brother October 18, 2013

2. ( )  2(-PE) m ( )  s 2(-(-8 (10-16) J)) m = 9.11 (10-31) kg Electron in TV Tube An electron is accelerated in a TV tube by a PD of 5000 V. find the potential energy change find the electron speed after accelerating EXAMPLE Solution (a) PE = qVba = (-1.6 (10-19) C) (5000 V) = - 8 (10-16) J, the potential energy has decreased (b) Wnet = -PE = KE = mv²− 0 v = = 4.2 (107)

3. X17 - 2 V 50 V = d 0.050 m V m Electric field obtained from voltage Two parallel plates are charged to 50 V. If the separation is 0.050 m find the field strength between them EXAMPLE Solution E = = 1000

4. Electric Potential of Point Charges k Q k Q 1 1 r1 r r² r2 kQq kQq r² r² ∫ dr -G M r The electric field strength is E = for a point charge Q The electric force is qE = The work done against the electric force when moving a charge is W = ∫ F·dr = = kQq− ( ) So the potential of a point charge Q is defined as V = This looks like the gravitational potential U = except that the electric potential depends on the sign of the charge

5. Electric Potential of Point Charges k Q k Q 1 1 r2 r r r1 V indicates the energy per positive unit charge to bring a charge close to Q V = when Q is positive If Q is + then you must do work to bring a + charge q close to it Q If Q is – then you get work from the + charge q as it approaches Q V = when Q is negative The work done against the electric force when moving a charge is W = kQq − ( )

6. X17 - 3 1 1 1 1 1 1 1 1 0.500 r2 0.04 r2 r1 0.02  r1 Nm² Nm² C² C² Work to force 2 +ve charges together Find the minimum work to bring 2 charges, + 3.00 C and + 20.00 C, to a distance of 0.500 m from a very large separation (r ~ ) EXAMPLE The work done against the electric force when moving a charge is W = kQq − Solution ( ) ( ) = 9 (109) (3.00 C) (20.0 C) − = 1.08 J EXAMPLE PE change when 2 charges separate What is the PE change when two -10 nC charges double their separation from 2 to 4 cm? The work done against the electric force when moving a charge is W = kQq− Solution ( ) ( ) = 9 (109) (-10 nC) (-10 nC) − = -22.5 (10-6) J

7. X2 - 3 Peer Instruction Will an electron move to a region of higher electric potential or lower electric potential? What about a proton? What happens to the potential energy in each case The potential is higher near a positive charge or region; electrons are attracted to positive charges, so an electron will move to higher potential – a proton will move to lower potential. In each case the potential energy decreases.

8. X2 - 3 Peer Instruction Can a particle move from a region of low electric potential to high electric potential and have its potential energy decrease? Yes; if the particle has a net negative charge on it – negative charges move naturally to positive (high potential) regions, accelerating and increasing kinetic energy while decreasing potential energy.

9. Problem solving with potentials A rQ rq q Q Potential is a scalar It is often easier to solve a problem using potentials rather than electric field strength which is a vector Total potential at a point is the sum of the potentials of all charges present VT = Σ V What is the potential at A due to Q and q? VA = VQA + VqA =kQ/rQ + kq/rq

10. k Q k q rQ rq A rQ rq q Q The Potential at a Point Due to 2 charges What is the potential at A due to Q and q? Let rq = rQ = 1 cm and q = 2 nC, Q = -4 nC EXAMPLE Solution Since total potential at a point is the sum of the potentials of all charges present we can just add VT = Σ V VA = VQA + VqA = + = − 1.8 kV

11. 3 μC -4 μC 0.2 m 0.20 m X 0.1 m 0.1 m -0.5 μC 2 μC EXAMPLE The Potential at a Point Find the potential at X Solution Since total potential at a point the sum of the potentials of all charges present we can add VT = Σ V = 90000 V

12. X2 - 3 Peer Instruction If the electric potential is 0 at a point does that mean the electric field is also 0 there? No; electric field is the change of electric potential divided by the distance between 2 points – the potential can go to 0 without the electric field being 0.

13. Q1 = 100 nC r1 = 5 cm X1 r = 20 cm kQ1 r1 X2 r2 = 5 cm kQ2 Q2 = -100 nC r + r2 EXAMPLE Accelerating a Charge 2 charges of ± 100 nC are shown; a 3rdcharge (10 nC, 5 μg) is released at X1; find its speed at X2 Find the potentials at X1 and X2 and the potential difference Then find the energy released (PE = W done) Solution From this find the speed V(X1) = + = 14,400 V V(X2) = -14,400 V V = V(X2) – V(X1) = -28,800 V W = -Vq = (-28,800 V)(10 (10-9)C) = 2.88 (10-4) J = KE v = 2W∕m = 340 m∕s

14. X2 - 3 Peer Instruction What can you say about the electric field in a region in which the potential is uniform or constant everywhere? Since electric field is the change of electric potential divided by the distance between 2 points if the potential does not change, the electric field must be 0

15. kQ1 kQ2 kQ2 kQ1 r2 r1 r1 r2 9(109)(-50(10-6)) 9(109)(50(10-6)) 0.60 0.30 The Potential Above 2 Point Charges Find the potentials at A and B EXAMPLE Solution (b) VB = Σ V (a) VA = Σ V = V1 + V2 = V1 + V2 = + = + = + = 0 V In fact everywhere on the y-axis the potential will be 0 = 1.5 (106) V – 0.75 (106) V = 7.5 (105) V

16. Required Before Next Class • Section 7.2 Review # 7, 8, 9 • Read Section 7.3