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Material Type: Exam; Class: General Physics II; Subject: Physics; University: La Salle University; Term: Spring 1994;
Typology: Exams
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84 A h 84 3600 3 0 105
C h s
s h
(b) The change in potential energy is ∆ U = q ∆ V = (3.0 × 105 C)(12 V) = 3.6 × 106 J.
the surface charge density. The field is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in the direction of the field. Then the electric potential is
V V (^) s E dx V Ex
x
where Vs is the potential at the sheet. The equipotential surfaces are surfaces of constant x ; that is, they are planes that are parallel to the plane of charge. If two surfaces are separated by ∆ x then
x
− −
12 2 6
C N m V C m
m
2 2
(b) VC – VA = VB – VA = 2.46 V.
(c) VC – VB = 0 (Since C and B are on the same equipotential line).
19 12 0 0 21 (^0 ) 0 0
f d i
q q d W q E ds dz
− − − −
is (in SI units)
12 2 12 0
z V
− − −
9 (^0 9 2 ) m /C
(10m) ( 1.0V) 4 1 8.99 10 N
⋅
( )
2 6 9 2 0 0 3
N m 1 1 1.0 10 C 8.99 10 4 4 C 2.0 m 1.0 m 4.5 10 V.
A B A B
q q V V
p p
(b) Since V ( r ) depends only on the magnitude of
r , the result is unchanged.
(a) For 0 < x < d we have d 1 = x and d 2 = d – x. Let
V x k
q d
q d
q x d x
F HG^
I KJ^
F H G
I K
(^1) J = 1
2 2 4 0
and solve: x = d /4. With d = 24.0 cm, we have x = 6.00 cm.
(b) Similarly, for x < 0 the separation between q 1 and a point on the x axis whose coordinate is x is given by d 1 = – x ; while the corresponding separation for q 2 is d 2 = d – x. We set
V x k q d
q d
q x d x
F H G
I K J =^ −
F H G
I K
(^1) J = 1
2 2 4 0
to obtain x = –d /2. With d = 24.0 cm, we have x = – 12.0 cm.
9 15 4 2 0 0
q q V
− − −
leftward, from the problem description (indicating deceleration of the rightward moving particle),
so that q > 0 (ensuring that F
→ is parallel to E
→ ); it is a proton.
(b) We use conservation of energy:
2 mpv
2 0 +^ qV 1 =
2 mpv
(^2) + qV
Using q = +1.6 × 10 −^19 C,^ m p = 1.67^ ×^10 −^27 kg,^ v 0 = 90^ ×^103 m/s,^ V 1 =^ −70 V and^ V 2^ =^ −50 V^ ,
we obtain the final speed v = 6.53 × 104 m/s. We note that the value of d is not used in the solution.
(b) Its final velocity, then, is in the negative x direction with a magnitude equal to that of its initial velocity. That is, its speed (upon leaving this region) is 1.0 × 107 m/s.
q = rV = × ⋅
m 1500 V N m C
b gb g
( )( ) 9 2 2 8 8 1 2 8.99^10 N m^ C^ 1.0^10 C^ 3.0^10 C^ 1.8 10 V.^2 4 2 1.0 m
q q V
p 0
(b) The distance from the center of one sphere to the surface of the other is d – R , where R is the radius of either sphere. The potential of either one of the spheres is due to the charge on that sphere and the charge on the other sphere. The potential at the surface of sphere 1 is
( )
8 8 1 2 9 2 2 3 1 0
8.99 10 N m C 2.9 10 V. 4 0.030 m 2.0 m 0.030 m
q q V
(c) The potential at the surface of sphere 2 is
( )
8 8 1 2 9 2 2 3 2 0
8.99 10 N m C 8.9 10 V. 4 2.0 m 0.030 m 0.030 m
q q V
1 2 m^ v
(^2) = e q 4 πεo r
where m = 9.11 × 10 −^31 kg, e = 1.60 × 10 −^19 C, q = 10000 e , and r = 0.010 m. This yields the answer v = 22490 m/s ≈ 2.2 × 104 m/s.
(a) Ex ( ab ) = –6.0 V/m,
(b) Ex ( bc ) = 0,
(c) Ex ( cd ) 3.0 V/m,
(d) Ex ( de ) = 3.0 V/m,
(e) Ex ( ef ) = 15 V/m,
(f) Ex ( fg ) = 0,