College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter18: Electric Charge And Electric Field
Section: Chapter Questions
Problem 25PE: Point charges of 5.00 C and 3.00/C are placed 0.250 m apart. (a) Where can a third charge be placed...
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Question
format in solving:
1. given
2. diagram (simple labeled sketch)
3. solution (use formula from formula sheet)
4. therefore statement (full answer with correct sig figs)
![FORMULA SHEET
Motion Equations:
Gravitational, Electric and Magnetic Fields:
d
Vave =
Vavg =
dr = d, + d2 + d3 + ……
ü = v, + ät
GMm
F =
GMm
E, =-
d, = d, + dz + dz+.
F, =
= ka
i =
+ ü)t
d = ö,t +
d = öt -
ü2 = v,? + 2äd
kog
E =
ΔΕ
AV =
V = kq
AE, =-AE
v²
a. =
q = Ne
= BIT sin 0
F = Bqü sin 0
HNI
Relative Motion:
B = ANI
Forces:
Modern Physics
Enet = må
= mỹ
F, = µFN
Fnet = F, + F2 + F,+...
Ato
At =
L = L. 1
m =
Energy and Momentum:
E = mc? – m.c?
E =
- moc²
p = mü
Ap = F At
= mAi
p = p'
Constants:
т — т,
g = 9.8 m/s² [Down]
k = 9.0 × 10° N - m² /C?
e =1.60 x 10-19C
H. = 47x10"T - m/ A
c = 3.0 x 10* m / s
v,'= v,|
т, +т,
= Amü
ElasticCollisions
G = 6.67 x 10-11 N-m²/kg²
2m,
v,'= v,
m, + m,
M, = 5.98×10ª kg
Earth
r = 6.37×10ʻm
m, =1.67×10" kg
m¸ = 9.11×10~" kg }Atom
r, = 5.29×10"m
E, -m
E. = kx?
W = F. Ad cos0
AE¸ = mgAh](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdee3419c-c208-4262-b310-a45b99a36ca0%2F5b4135c9-2309-4661-9313-84d1dc007aab%2Fbn24ikn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:FORMULA SHEET
Motion Equations:
Gravitational, Electric and Magnetic Fields:
d
Vave =
Vavg =
dr = d, + d2 + d3 + ……
ü = v, + ät
GMm
F =
GMm
E, =-
d, = d, + dz + dz+.
F, =
= ka
i =
+ ü)t
d = ö,t +
d = öt -
ü2 = v,? + 2äd
kog
E =
ΔΕ
AV =
V = kq
AE, =-AE
v²
a. =
q = Ne
= BIT sin 0
F = Bqü sin 0
HNI
Relative Motion:
B = ANI
Forces:
Modern Physics
Enet = må
= mỹ
F, = µFN
Fnet = F, + F2 + F,+...
Ato
At =
L = L. 1
m =
Energy and Momentum:
E = mc? – m.c?
E =
- moc²
p = mü
Ap = F At
= mAi
p = p'
Constants:
т — т,
g = 9.8 m/s² [Down]
k = 9.0 × 10° N - m² /C?
e =1.60 x 10-19C
H. = 47x10"T - m/ A
c = 3.0 x 10* m / s
v,'= v,|
т, +т,
= Amü
ElasticCollisions
G = 6.67 x 10-11 N-m²/kg²
2m,
v,'= v,
m, + m,
M, = 5.98×10ª kg
Earth
r = 6.37×10ʻm
m, =1.67×10" kg
m¸ = 9.11×10~" kg }Atom
r, = 5.29×10"m
E, -m
E. = kx?
W = F. Ad cos0
AE¸ = mgAh
Transcribed Image Text:Locate the point at which a positive charge can be placed and experience a zero net
force.
-5.00 C
- 1.75C
92
0.500 m
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