Review for the Chapter test Page 1 Page 2 Page 3 Page 4

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

+.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

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)(1.602x C) = 5.21x N up F = ma, a = F/m = (5.21x N)/(9.11x kg) = 5.72x10 13 m/s/s

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 = C)

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

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 = N/C (to the right) r 2 E B = kq B = N/C (to the right) r 2 = N/C right N/C right = N/C right = 2.7E6 N/C right x

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  C-4.10  C q 12.0 cm  C Initial V V Final V V Change in V V Work4.448 V +4.4 J

Page 2 toc

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?

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

Which way is the electric field? (wwpcd?)

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?

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

.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

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 = V 3.  V = V - 0 = V 4.W = Vq = ( J/C)(6E-6C) = J

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  C-4.10  C Q 12.0 cm  C Initial V V {k(3.2E-6)/.12 + k(-4.10E-6)/.12} k = 8.99E9 Final V V {k(3.2E-6)/.06 + k(-4.10E-6)/.18} Change in V V {Final - initial} Work4.448 V {W =  Vq, q =  C - the moved charge } +4.4 J

Page 3 toc

.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 kg

9.61x N W E = F/q, E =.0600 V/m, q = x C F = x N = 9.61x 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 kg

1.06x10 10 m/s/s W F = ma, F = x N, m = 9.11x kg a = x10 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 kg

3.92x10 -6 s W V = s/t, V = 114,700, s =.45 m t = x10 -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 kg

8.12 cm W s = ut + 1 / 2 at 2, u = 0, t = x10 -6 s, a = x10 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 kg

.0374 V W Vq = 1 / 2 mv 2, q = 1.602x C, v = 114,700, m = 9.11x 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 kg

Page 4 toc

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 = N/C (to the right) r 2 E B = kq B = N/C (to the right) r 2 = N/C right N/C right = N/C right = 2.7E6 N/C right x

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 = N, F BC = N  ABC = Tan -1 (.92/1.9) = o F AC = 0 N x N y F BC = cos(25.84 o ) x sin(25.84 o )y F total = x+ 369 y 410 N, 65o above x axis (to the left of y)

TOC Q2Q2 Q1Q  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  ( ) Voltage due to Q 2 : V 2 = kq 2 =k(3.1x10 -6 )= 1.36x10 4 V r  ( ) + 2.6x10 4 V And The Sum Is…

-14,000 V W V 1 = V 2 = V 1 + V 1 = = -14,000 V Find the voltage at point C Q2Q2 Q1Q  C +1.1  C 38 cm 85 cm C

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 = V 3.  V = V - 0 = V 4.W = Vq = ( J/C)(6E-6C) = J

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  C-4.10  C Q 12.0 cm  C Initial V V {k(3.2E-6)/.12 + k(-4.10E-6)/.12} k = 8.99E9 Final V V {k(3.2E-6)/.06 + k(-4.10E-6)/.18} Change in V V {Final - initial} Work4.448 V {W =  Vq, q =  C - the moved charge } +4.4 J