Unformatted text preview: e point charge should have the same value Solutions will be posted at www.prep101.com/solutions 13 ©Prep101 Phys1026 April Exam Solutions qpoint = 7.85 x107 C
b) This point is inside a conductor – the electric field is always zero. The minimum
point charge required is zero! ELECTRIC POTENTIAL
PRACTICE QUESTIONS
Question 1:
Solution:
The Work done on the electron by the electric field is given by the equation: ( )( W = q Δ V = 1 .6 x10 − 19 3.0x10 3 ) = 4 .8 x10 − 16 J Since the object starts from rest, we can assume that all of this work is converted to
1
kinetic energy of the electron. Thus, we get K = me v 2 = 4.8 x10 −16 J
2
Solving for v we get: v= ( ) 2 4.8 x10 −16
= 3.2 x10 7 m s
−31
9.11x10 Question 2:
Solution: K= 12
6.7 × 10 −22 (4.8 × 10 5 ) 2
mv 2
v
kv
mv = ΔU = −qEΔx ⇔ E =
=
= 1206 = 1.2
−19
m
m
2
− 2qΔx − 2 ⋅ 3.2 ⋅ 10 ⋅ (−2m) Question 3:
Solution:
For A: U = q2V1 = k e q 1q 2 (8.99 ⋅ 10 9 Nm 2 /C 2 )(100 ⋅ 10 −6 C)(−300 ⋅ 10 −6 C)
=
r12
0.3 m U = 899 J
For B: Vp = ⎛ 100 ⋅ 10 −6 C − 300 ⋅ 10 −6 C ⎞
kq 1 kq 2
⎟
+
= (8.99 ⋅ 10 9 Nm 2 /C 2 )⎜
+
⎜
⎟
r1
r2
1m
0.8 m
⎝
⎠ Solutions will be posted at www.prep101.com/solutions 14 ©Prep101 Phys1026 April Exam Solutions Vp = 2.47⋅106 V For C:
U = eVP
U = (1.602⋅1019 C) ⋅ (2.47⋅106 V)
U = 3.96⋅1013 J
Question 4:
Solution:
For A:
r
kq 1q 2 (8.99 ⋅ 10 9 )(10 ⋅ 10 −9 )(100 ⋅ 10 −9 )
F=
= 8.99x106 N
=
2
2
r
(1) For B:
Yes, since the charges have opposite signs. For C:
vv
v
E = E1 + E 2 E= kq 1
r1P 2 + kq 2
r2P 2 = (8.99 ⋅ 10 9 Nm 2 / C 2 ) ⋅ (110 ⋅ 10 −9 C)
(0.5m) 2 E = 3956 N/C; where the direction of the electric field is to the right
For D: V=− kq 2
(8.99 ⋅ 10 9 Nm 2 / C 2 ) ⋅ (100 ⋅ 10 −9 C)
=−
= 899 V
r12
(1 m) Solutions will be posted at www.prep101.com/solutions 15 ©Prep101 Phys1026 April Exam Solutions Question 5:
Solution:
For A: V = kq / r = +900 V
q = 900 r / k q= (0.07 m) (900 V)
(9 x10 9 Nm 2 /C 2 ) q = +7x10−9 Coulombs
For B:
Any metallic solid has a constant potential throughout. Therefore, the potential difference
between the surface and the center = 0 V
For C:
For r > 7 cm: V(r) = kq / r (9 ⋅ 10 9 )(7 x10 −9 )
V(21 cm) =
= +300 V
0.21
At r = 7 cm: V = 900 V ΔV = V7 – V21 = 900 V – 300 V = +600 V
For D: New charge q ′ = Vr (909)(0.07 )
=
= 7.07 x10 −9 C
9
k
9x10 ∴Δq = q′  qinit = (7.07 – 7) x 10−9 Coulombs
∴ΔU = ΔV Δq = (900 V) ⋅ (0.07x109 C) = 6.3x108 J
Question 6: Solution:
First, let’s call to the right to be the positive direction
For A:
Since A is negative, the electric field at P due to A will point toward A (to the left). Since
B is positive, the electric field at P due to B will point away from B (to the left as well).
−6
9 N ⋅m
kQ A 9.00 × 10 c 2 (−2.50 × 10 C )
= −9.00 × 10 4
EA = 2 =
2
rA
(0.500m )
2 N C −6
9 N ⋅m
kQB 9.00 × 10 c 2 (5.00 × 10 C )
= 1.80 × 10 5 N C
EB = 2 =
2
rB
(0.500m )
2 Solutions will be posted at www.prep1...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.
 Winter '14
 Prof.PaulineBarmby
 Physics

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