2. Review for the Chapter 16-17 test Page 1 Page 2 Page 3 Page 4

3. Page 1 These examples all have to do with electric field – the question on the test has to do with gravity, but parallels these questions. Best also look at the suggested review questions on the Field Theory worksheet toc

4. +.11  C E for a point charge: E = kq r 2 k = 8.99x10 9 Nm 2 C -2, E = 2,120 N/C, r =.67 m q = 1.06x10 -7 C = +.11  C. It is a positive charge as the E-field is away from it Vesta Buhl measures an electric field of 2,120 N/C, 67 cm from a charge of unknown value. The electric field is away from the charge. What is the charge? W

5. Electric Field TOC Example 2 - An electron travels through a region where there is a downward electric field of 325 N/C. What force in what direction acts on the electron, and what is its acceleration? F = Eq = (325 N/C)(1.602x10 -19 C) = 5.21x10 -17 N up F = ma, a = F/m = (5.21x10 -17 N)/(9.11x10 -31 kg) = 5.72x10 13 m/s/s

6. 11,000 N W F = kq 1 q 2 r 2 k = 8.99x10 9 Nm 2 C -2, q 1 = 3.0x10 -3 C, q 2 = 5.0 x10 -3 C, r = 3.5 m F = 11,000 N Jess Uwaite places a +3.0 mC charge 3.5 m from a +5.0 mC charge. What is the force of repulsion? (1 mC = 10 -3 C)

7. 180 km W Noah Verkreinatlaad places a 5.0 C charge how far from a 3.0 C charge to make the force between them exactly 4.00 N? F = kq 1 q 2 r 2 k = 8.99x10 9 Nm 2 C -2, q 1 = 5.0 C, q 2 = 3.0 C, F = 4.0 N r = 1.8x10 5 m = 180 km = 100 miles wow

8. 41 N left W What is the electric field at the x? Which Direction is it? AB +120  C -180  C 70. cm 170 cm E A = kq A = 2201632.653 N/C (to the right) r 2 E B = kq B = 559930.7958 N/C (to the right) r 2 = 2201632.653 N/C right + 559930.7958 N/C right = 2761563.449 N/C right = 2.7E6 N/C right x

9. Try this one TOC What work to bring a 13.0  C charge from halfway between the other two charges to 6.0 cm from the positive and 18 cm from the negative? qq +3.20  C-4.10  C q 12.0 cm +13.0  C Initial V-67425 V Final V274700. V Change in V342100. V Work4.448 V +4.4 J

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11. 5.0x10 2 V/m W E = V/d, V = 25, d =.050 m E = 500 V/m = 5.0x10 2 V/m Lee DerHosen places a voltage of 25 V across two || plates separated by 5.0 cm of distance. What is the electric field generated?

12. Electric Field TOC Example 1 - A +125  C charge experiences a force to the right of.0175 N. What is the Electric field, and its direction? E = F/q =.0175 N/125x10 -6 C = 140 N/C to the right E Direction: +Q Force This Way -Q Force This Way

13. ++++++++++ ---------- Which way is the electric field? (wwpcd?)

14. 10.0V W V = W/q, W = 125 J, q = 12.5 C V = 10.0 V Sandy Deck does 125 J of work on a 12.5 C charge. Through what voltage did she move it?

15. 726,000 m/s W V = W/q, W = Vq = 1 / 2 mv 2 V = 1.50 V, m = 9.11x10 -31 kg, q = 1.602x10 -19 C v = 726327.8464 = 726,000 m/s Brennan Dondahaus accelerates an electron (m = 9.11x10 -31 kg) through a voltage of 1.50 V. What is its final speed assuming it started from rest?

16. .22  C W Alex Tudance measures a voltage of 25,000 volts near a Van de Graaff generator whose dome is 7.8 cm in radius. What is the charge on the dome? V = kq/r, r =.078 m, V = 25,000 V q = 2.17x10 -7 C =.22  C

17. Cute problems with voltage TOC What work to bring a 6  C charge from infinity to halfway between the other two charges? QQ +1.5  C 24.0 cm 1.Find initial voltage = 0 (at infinity) 2.Find final voltage = k(1.5E-6)/.12 + k(1.5E-6)/.12 = 224750 V 3.  V = 224750 V - 0 = 224750 V 4.W = Vq = (224750 J/C)(6E-6C) = 1.3485 J

18. Try this one TOC What work to bring a 13.0  C charge from halfway between the other two charges to 6.0 cm from the positive and 18 cm from the negative? QQ +3.20  C-4.10  C Q 12.0 cm +13.0  C Initial V-67425 V {k(3.2E-6)/.12 + k(-4.10E-6)/.12} k = 8.99E9 Final V274700. V {k(3.2E-6)/.06 + k(-4.10E-6)/.18} Change in V342100. V {Final - initial} Work4.448 V {W =  Vq, q = +13.0  C - the moved charge } +4.4 J

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20. .0600 V/m W E = V/d, V =.0120 V, d =.200 m E =.0600 V/m An electron traveling 114,700 m/s parallel to the plates above, and midway between them is deflected upward by a potential of.0120 V. A. What is the electric field between the plates? 45.0 cm 20.0 cm me = 9.11 x 10 -31 kg

21. 9.61x10 -21 N W E = F/q, E =.0600 V/m, q = -1.602x10 -19 C F = 9.6120x10 -21 N = 9.61x10 -21 N An electron traveling 114,700 m/s parallel to the plates above, and midway between them is deflected upward by a potential of.0120 V. B. What is the electrical force on the electron between the plates? 45.0 cm 20.0 cm me = 9.11 x 10 -31 kg

22. 1.06x10 10 m/s/s W F = ma, F = 9.6120x10 -21 N, m = 9.11x10 -31 kg a = 1.0551x10 10 m/s/s = 1.06x10 10 m/s/s (You can neglect gravity) An electron traveling 114,700 m/s parallel to the plates above, and midway between them is deflected upward by a potential of.0120 V. C. What is the upward acceleration of the electron between the plates? 45.0 cm 20.0 cm me = 9.11 x 10 -31 kg

23. 3.92x10 -6 s W V = s/t, V = 114,700, s =.45 m t = 3.9233x10 -6 s = 3.92x10 -6 s An electron traveling 114,700 m/s parallel to the plates above, and midway between them is deflected upward by a potential of.0120 V. D. For what time is the electron between the plates? 45.0 cm 20.0 cm me = 9.11 x 10 -31 kg

24. 8.12 cm W s = ut + 1 / 2 at 2, u = 0, t = 3.9233x10 -6 s, a = 1.0551x10 10 m/s/s s =.0812 m = 8.12 cm An electron traveling 114,700 m/s parallel to the plates above, and midway between them is deflected upward by a potential of.0120 V. E. What is the vertical displacement of the electron while is passes between the plates? 45.0 cm 20.0 cm me = 9.11 x 10 -31 kg

25. .0374 V W Vq = 1 / 2 mv 2, q = 1.602x10 -19 C, v = 114,700, m = 9.11x10 -31 kg V =.0374 V An electron traveling 114,700 m/s parallel to the plates above, and midway between them is deflected upward by a potential of.0120 V. F. Through what potential was the electron accelerated to reach a velocity of 114,700 m/s from rest? 45.0 cm 20.0 cm me = 9.11 x 10 -31 kg

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27. 41 N left W What is the electric field at the x? Which Direction is it? AB +120  C -180  C 70. cm 170 cm E A = kq A = 2201632.653 N/C (to the right) r 2 E B = kq B = 559930.7958 N/C (to the right) r 2 = 2201632.653 N/C right + 559930.7958 N/C right = 2761563.449 N/C right = 2.7E6 N/C right x

28. W ABC +180  C +150  C +520  C 1.9 m.92 m Find the force on C, and the angle it makes with the horizontal. (the one on the test is electric field…) F AC = 286.8 N, F BC = 188.8 N  ABC = Tan -1 (.92/1.9) = 25.84 o F AC = 0 N x+ 286.8 N y F BC = -188.8cos(25.84 o ) x+ 188.8sin(25.84 o )y F total = -170. x+ 369 y 410 N, 65o above x axis (to the left of y)

29. TOC Q2Q2 Q1Q1 +1.5  C +3.1  C 190 cm75 cm Find the voltage at point A: A Voltage at A is scalar sum of V 1 and V 2 : Voltage due to Q 1 : V 1 = kq 1 =k(1.5x10 -6 )= 1.27x10 4 V r  (.75 2 +.75 2 ) Voltage due to Q 2 : V 2 = kq 2 =k(3.1x10 -6 )= 1.36x10 4 V r  (.75 2 +1.9 2 ) + 2.6x10 4 V And The Sum Is…

30. -14,000 V W V 1 =-39587.58847 V 2 =+26023.68421 V 1 + V 1 =-13563.90426 = -14,000 V Find the voltage at point C Q2Q2 Q1Q1 -4.1  C +1.1  C 38 cm 85 cm C

31. Cute problems with voltage TOC What work to bring a 6  C charge from infinity to halfway between the other two charges? QQ +1.5  C 24.0 cm 1.Find initial voltage = 0 (at infinity) 2.Find final voltage = k(1.5E-6)/.12 + k(1.5E-6)/.12 = 224750 V 3.  V = 224750 V - 0 = 224750 V 4.W = Vq = (224750 J/C)(6E-6C) = 1.3485 J

32. Try this one TOC What work to bring a 13.0  C charge from halfway between the other two charges to 6.0 cm from the positive and 18 cm from the negative? QQ +3.20  C-4.10  C Q 12.0 cm +13.0  C Initial V-67425 V {k(3.2E-6)/.12 + k(-4.10E-6)/.12} k = 8.99E9 Final V274700. V {k(3.2E-6)/.06 + k(-4.10E-6)/.18} Change in V342100. V {Final - initial} Work4.448 V {W =  Vq, q = +13.0  C - the moved charge } +4.4 J