Transcript Review for 16-17x - TuHS Physics Homepage

Review for the Chapter 16-17 test
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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
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?
E for a point charge:
E = kq
r2
k = 8.99x109 Nm2C-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
+.11 C
W
Electric Field
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)/(9.11x10-31kg) = 5.72x1013 m/s/s
TOC
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)
F = kq1q2
r2
k = 8.99x109 Nm2C-2, q1 = 3.0x10-3 C, q2 = 5.0 x10-3 C,
r = 3.5 m
F = 11,000 N
11,000 N
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 N
r = 1.8x105 m = 180 km = 100 miles wow
180 km
W
-180 C
+120 C
A
70. cm
170 cm
B
x
What is the electric field at the x? Which
Direction is it?
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
41 N left
W
Try this one
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?
+3.20 C
+13.0 C
-4.10 C
12.0 cm
12.0 cm
q
q
q
Initial V
Final V
Change in V
Work
+4.4 J
-67425 V
274700. V
342100. V
4.448 V
TOC
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Lee DerHosen places a voltage of 25 V
across two || plates separated by 5.0
cm of distance. What is the electric
field generated?
E = V/d, V = 25, d = .050 m
E = 500 V/m = 5.0x102 V/m
5.0x102 V/m
W
Electric Field
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
Direction:
+Q
Force This Way
E
Force This Way
-Q
TOC
Which way is the electric field? (wwpcd?)
+
+
+
+
+
-
Sandy Deck does 125 J of work on a
12.5 C charge. Through what voltage
did she move it?
V = W/q, W = 125 J, q = 12.5 C
V = 10.0 V
10.0V
W
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?
V = W/q, W = Vq = 1/2mv2
V = 1.50 V, m = 9.11x10-31 kg, q = 1.602x10-19 C
v = 726327.8464 = 726,000 m/s
726,000 m/s
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
.22 C
W
Cute problems with voltage
What work to bring a 6 C charge from infinity to halfway
between the other two charges?
+1.5 C
Q
1.
2.
3.
4.
+1.5 C
24.0 cm
Q
Find initial voltage = 0 (at infinity)
Find final voltage = k(1.5E-6)/.12 + k(1.5E-6)/.12 = 224750 V
V = 224750 V - 0 = 224750 V
W = Vq = (224750 J/C)(6E-6C) = 1.3485 J
TOC
Try this one
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?
+3.20 C
+13.0 C
-4.10 C
12.0 cm
12.0 cm
Q
Q
Q
Initial V
Final V
Change in V
Work
+4.4 J
-67425 V
274700. V
342100. V
4.448 V
{k(3.2E-6)/.12 + k(-4.10E-6)/.12} k = 8.99E9
{k(3.2E-6)/.06 + k(-4.10E-6)/.18}
{Final - initial}
{W = Vq, q = +13.0 C - the moved charge}
TOC
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toc
45.0 cm
me = 9.11 x 10-31 kg
20.0 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.
A. What is the electric field between the plates?
E = V/d, V = .0120 V, d = .200 m
E = .0600 V/m
.0600 V/m
W
45.0 cm
me = 9.11 x 10-31 kg
20.0 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.
B. What is the electrical force on the electron
between the plates?
E = F/q, E = .0600 V/m, q = -1.602x10-19 C
F = 9.6120x10-21 N = 9.61x10-21 N
9.61x10-21 N
W
45.0 cm
me = 9.11 x 10-31 kg
20.0 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.
C. What is the upward acceleration of the electron
between the plates?
F = ma, F = 9.6120x10-21 N, m = 9.11x10-31 kg
a = 1.0551x1010 m/s/s = 1.06x1010 m/s/s
(You can neglect gravity)
1.06x1010 m/s/s
W
45.0 cm
me = 9.11 x 10-31 kg
20.0 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.
D. For what time is the electron between the plates?
V = s/t, V = 114,700, s = .45 m
t = 3.9233x10-6 s = 3.92x10-6 s
3.92x10-6 s
W
45.0 cm
me = 9.11 x 10-31 kg
20.0 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?
s = ut + 1/2at2, u = 0, t = 3.9233x10-6 s, a = 1.0551x1010 m/s/s
s = .0812 m = 8.12 cm
8.12 cm
W
45.0 cm
me = 9.11 x 10-31 kg
20.0 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.
F. Through what potential was the electron accelerated to
reach a velocity of 114,700 m/s from rest?
Vq = 1/2mv2, q = 1.602x10-19 C, v = 114,700, m = 9.11x10-31 kg
V = .0374 V
.0374 V
W
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toc
-180 C
+120 C
A
70. cm
170 cm
B
x
What is the electric field at the x? Which
Direction is it?
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
41 N left
W
C
+180 C
.92 m
+150 C
A
Find the force on C, and
the angle it makes with the
horizontal. (the one on the
test is electric field…)
1.9 m
+520 C
B
FAC= 286.8 N, FBC = 188.8 N
ABC = Tan-1(.92/1.9) = 25.84o
FAC = 0 N x
+ 286.8 N y
FBC = -188.8cos(25.84o) x + 188.8sin(25.84o)y
Ftotal =
-170. x
+ 369 y
410 N, 65o above x axis (to the left of y)
W
A
Find the voltage at point A:
+1.5 C
Q1
75 cm
+3.1 C
190 cm
Q2
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…
TOC
Find the voltage at point C
-4.1 C
Q1
V1
=
V2
=
V1 + V1 =
-14,000 V
C
38 cm
85 cm
Q2
+1.1 C
-39587.58847
+26023.68421
-13563.90426 = -14,000 V
W
Cute problems with voltage
What work to bring a 6 C charge from infinity to halfway
between the other two charges?
+1.5 C
Q
1.
2.
3.
4.
+1.5 C
24.0 cm
Q
Find initial voltage = 0 (at infinity)
Find final voltage = k(1.5E-6)/.12 + k(1.5E-6)/.12 = 224750 V
V = 224750 V - 0 = 224750 V
W = Vq = (224750 J/C)(6E-6C) = 1.3485 J
TOC
Try this one
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?
+3.20 C
+13.0 C
-4.10 C
12.0 cm
12.0 cm
Q
Q
Q
Initial V
Final V
Change in V
Work
+4.4 J
-67425 V
274700. V
342100. V
4.448 V
{k(3.2E-6)/.12 + k(-4.10E-6)/.12} k = 8.99E9
{k(3.2E-6)/.06 + k(-4.10E-6)/.18}
{Final - initial}
{W = Vq, q = +13.0 C - the moved charge}
TOC