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

Review for the Chapter 16-17 test

Page 1Page 2Page 3Page 4

Page 1These 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

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+.11 C

E for a point charge:E = kq

r2

k = 8.99x109 Nm2C-2, E = 2,120 N/C, r = .67 mq = 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

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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 upF = ma, a = F/m = (5.21x10-17 N)/(9.11x10-31kg) = 5.72x1013 m/s/s

11,000 N W

F = kq1q2

r2

k = 8.99x109 Nm2C-2, q1 = 3.0x10-3 C, q2 = 5.0 x10-3 C, r = 3.5 mF = 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)

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 = kq1q2

r2

k = 8.99x109 Nm2C-2, q1 = 5.0 C, q2 = 3.0 C, F = 4.0 Nr = 1.8x105 m = 180 km = 100 miles wow

41 N left W

What is the electric field at the x? Which Direction is it?

A B

+120 C -180 C

70. cm 170 cm

EA = kqA = 2201632.653 N/C (to the right)

r2

EB = kqB = 559930.7958 N/C (to the right)

r2

= 2201632.653 N/C right + 559930.7958 N/C right = 2761563.449 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

+3.20 C -4.10 C

q12.0 cm12.0 cm

+13.0 C

Initial V -67425 VFinal V 274700. VChange in V 342100. VWork 4.448 V

+4.4 J

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5.0x102 V/m W

E = V/d, V = 25, d = .050 mE = 500 V/m = 5.0x102 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

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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:

+QForce This Way

-QForce This Way

+++++

-----

Which way is the electric field? (wwpcd?)

10.0V W

V = W/q, W = 125 J, q = 12.5 CV = 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 = 1/2mv2

V = 1.50 V, m = 9.11x10-31 kg, q = 1.602x10-19 Cv = 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?

.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 Vq = 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 +1.5 C24.0 cm

1. Find initial voltage = 0 (at infinity)2. Find final voltage = k(1.5E-6)/.12 + k(1.5E-6)/.12 = 224750 V3. V = 224750 V - 0 = 224750 V4. W = Vq = (224750 J/C)(6E-6C) = 1.3485 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

+3.20 C -4.10 C

Q12.0 cm12.0 cm

+13.0 C

Initial V -67425 V {k(3.2E-6)/.12 + k(-4.10E-6)/.12} k = 8.99E9Final V 274700. V {k(3.2E-6)/.06 + k(-4.10E-6)/.18}Change in V 342100. V {Final - initial}Work 4.448 V {W = Vq, q = +13.0 C - the moved charge}

+4.4 J

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.0600 V/m W

E = V/d, V = .0120 V, d = .200 mE = .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

9.61x10-21 N W

E = F/q, E = .0600 V/m, q = -1.602x10-19 CF = 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

1.06x1010 m/s/s W

F = ma, F = 9.6120x10-21 N, m = 9.11x10-31 kga = 1.0551x1010 m/s/s = 1.06x1010 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

3.92x10-6 s W

V = s/t, V = 114,700, s = .45 mt = 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

8.12 cm W

s = ut + 1/2at2, u = 0, t = 3.9233x10-6 s, a = 1.0551x1010 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

.0374 V W

Vq = 1/2mv2, 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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41 N left W

What is the electric field at the x? Which Direction is it?

A B

+120 C -180 C

70. cm 170 cm

EA = kqA = 2201632.653 N/C (to the right)

r2

EB = kqB = 559930.7958 N/C (to the right)

r2

= 2201632.653 N/C right + 559930.7958 N/C right = 2761563.449 N/C right = 2.7E6 N/C right

x

W

A B

C +180 C

+150 C +520 C1.9 m

.92 m

Find the force on C, and the angle it makes with the horizontal. (the one on the test is electric field…)

FAC= 286.8 N, FBC = 188.8 NABC = Tan-1(.92/1.9) = 25.84o

FAC = 0 N x + 286.8 N yFBC = -188.8cos(25.84o) x + 188.8sin(25.84o)y

Ftotal = -170. x + 369 y410 N, 65o above x axis (to the left of y)

TOC

Q2Q1

+1.5 C +3.1 C190 cm75 cm

Find the voltage at point A:A

Voltage at A is scalar sum of V1 and V2:Voltage due to Q1:

V1 = kq1 = k(1.5x10-6) = 1.27x104 V r (.752+.752)

Voltage due to Q2:V2 = kq2 = k(3.1x10-6) = 1.36x104 V

r (.752+1.92)

+ 2.6x104 V

And The Sum Is…

-14,000 V W

V1 = -39587.58847V2 = +26023.68421V1 + V1 = -13563.90426 = -14,000 V

Find the voltage at point C

Q2Q1

-4.1 C

+1.1 C

38 cm85 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 +1.5 C24.0 cm

1. Find initial voltage = 0 (at infinity)2. Find final voltage = k(1.5E-6)/.12 + k(1.5E-6)/.12 = 224750 V3. V = 224750 V - 0 = 224750 V4. W = Vq = (224750 J/C)(6E-6C) = 1.3485 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

+3.20 C -4.10 C

Q12.0 cm12.0 cm

+13.0 C

Initial V -67425 V {k(3.2E-6)/.12 + k(-4.10E-6)/.12} k = 8.99E9Final V 274700. V {k(3.2E-6)/.06 + k(-4.10E-6)/.18}Change in V 342100. V {Final - initial}Work 4.448 V {W = Vq, q = +13.0 C - the moved charge}

+4.4 J