Fisica Cap 23 Tippens

Chapter 25. Electric Potential

Work and Electric Potential Energy

25-1. An positively charged plate is 30 mm above a negatively charged plate, and the electric field intensity has a magnitude of 6 x 104 N/C. How much work is done BY the electric field when a +4-C charge is moved from the negative plate to the positive plate?

Work = Fd = qEd; F opposite to d makes work negative.

Work = (4 x 10-6 C)(6 x 104 N/C)(0.030 m)

Work = -7.20 x 10-3 J; Work = -7.20 mJ

25-2. In Problem 25-1, how much work is done ON or against the electric field? What is the electric potential energy at the positive plate?

The displacement occurs against the electric force, so that

work done ON the field due to outside UP force Fext in same direction as the displacement.

Since the field is in a position to do positive work when at the positive, plate, the electric potential energy is positive at that point: Ep = +7.20 mJ.

25-3. The electric field intensity between two parallel plates, separated by 25 mm is 8000 N/C. How much work is done BY the electric field in moving a –2-C charge from the negative plate to the positive plate? What is the work done BY the field in moving the same charge back to the positive plate? (Electric force with motion)

Work done BY field is positive, Fe with displacement.

Work = qEd = (2 x 10-6 C)(8000 N/C)(0.025 m) Work done BY = Loss of electric energy.

Work = 4.00 x 10-4 J Now, in coming back electric force opposes motion.

Work done BY field is negative: Work = - 4.00 x 10-4 J.

25-4. In Problem 25-3, what is the potential energy when the charge is at (a) the positive plate and (b) the negative plate?

Remember, potential energy represents the work that the

electric field can do when electric forces are free to act. When the –2 C charge is at the positive plate, the E field can do no work, thus with reference to that point, Ep = 0.

(a) At + plate: Ep = 0 No work can be done by the electric field.

(b) At – plate, the field can do +work: Ep = +4.00 x 10-4 J

25-5. What is the potential energy of a +6 nC charge located 50 mm away from a +80-C charge? What is the potential energy if the same charge is 50 mm from a –80-C charge?

P.E. = 86.4 mJ

; P.E. = -86.4 mJ

25-6. At what distance from a –7-C charge will a –3-nC charge have a potential energy of 60 mJ? What initial force will the –3-nC charge experience?

; r = 3.15 mm

; F = 19.0 N, repulsion

Note: This value can also be obtained from:

25-7. A +8-nC charge is placed at a point P, 40 mm from a +12-C charge. What is the potential energy per unit charge at point P in joules per coulomb? Will this change if the 8-nC charge is removed?

P.E. = 0.0216 J;

V = 2.70 x 106 J/C ; No The P.E./q is a property of space. If another charge were placed there or if no charge were there, the P.E./q is the same.

25-8. A charge of +6 C is 30 mm away from another charge of 16 C. What is the potential energy of the system?

; P.E. = 28.8 J

25-9. In Problem 25-8, what is the change in potential energy if the 6-C charge is moved to a distance of only 5 mm? Is this an increase or decrease in potential energy?

(P.E.)30 = 28.8 J from previous example. Now assume charge is moved.

; (P.E.)5 = 173 J

Change in P.E. = 172. J – 28.8 J; Change = 144 J, increase

25-10. A –3-C charge is placed 6 mm away from a -9-C charge. What is the potential energy? Is it negative or positive?

; P.E. = +40.5 J

25-11. What is the change in potential energy when a 3-nC charge is moved from a point 8 cm away from a –6-C charge to a point that is 20 cm away? Is this an increase or decrease of potential energy? (Moves from A to B on figure.)

(P.E.)8 = -2.025 J, (Negative potential energy)

(P.E.)20 = -0.810 J,

Change = final – initial = -0.810 J – (-2.025 J); Change in P.E. = +1.22 J, increase

25-12. At what distance from a –7-C charge must a charge of –12 nC be placed if the potential energy is to be 9 x 10-5 J?

; r = 8.40 m

25-13. The potential energy of a system consisting of two identical charges is 4.50 mJ when their separation is 38 mm. What is the magnitude of each charge?

; q = 138 nC

Electric Potential and Potential Difference

25-14. What is the electric potential at a point that is 6 cm from a 8.40-C charge? What is the potential energy of a 2 nC charge placed at that point?

; V = 1.26 x 106 V

P.E. = qV = (2 x 10-9 C)(1.26 x 106 V); P.E. = 2.52 mJ

25-15. Calculate the potential at point A that is 50 mm from a –40-C charge. What is the potential energy if a +3-C charge is placed at point A?

; V = -7.20 x 106 V

P.E. = qV = (3 x 10-6 C)(-7.2 x 106); P.E. = -21.6 J

25-16. What is the potential at the midpoint of a line joining a –12-C charge with a +3-C charge located 80 mm away from the first charge?

(Net potential is algebraic sum)

V = -2.70 x 106 V + 0.675 x 106 V; V = -2.025 x 106 V; V = -2.02 MV

25-17. A +45-nC charge is 68 mm to the left of a –9-nC charge. What is the potential at a point located 40 mm to the left of the –9-nC charge?

Find potential due to each charge, then add:

V = +14.5 x 103 V + (-2.025 x 103 V); V = +12.4 kV

*25-18. Points A and B are 68 mm and 26 mm away from a 90-C charge. Calculate the potential difference between points A and B? How much work is done BY the electric field as a -5-C charge moves from A to B?

; VB = 3.115 x 107 V

*25-18. (Cont.) ;

VA = 1.19 x 107 V; VB = 3.115 x 107 V

VB – VA = 3.115 x 107 V – 1.19 x 107 V; V = 1.92 x 107 V

Note that the potential INCREASES because B is at a higher potential than A

Now for the field: (Work)AB = q(VA - VB) = (-5 x 10-6 C)(1.19 x 107 V – 3.119 x 107 V);

WorkAB = +96.2 mJ; The field does positive work on a negative charge.

*25-19. Points A and B are 40 mm and 25 mm away from a +6-C charge. How much work must be done against the electric field (by external

...