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Physics 2020: Sample Problems for Exam 1
- Two particles are held fixed on the x-axis. The first particle has a charge of Q 1 = 6. 88 × 10 −^5 C and is located at x 1 = − 4 .56 m on the x-axis. The second particle has a charge of Q 2 = 2. 33 × 10 −^4 C and is located at x 2 = 8.05 m on the x-axis. A third particle of charge q < 0 and mass m = 0.765 g is located at +2.89 m on the x-axis and is free to move. (a) If charge q experiences a net electric force of +5. 56 × 10 −^4 N at this location from the fixed charges, what is the charge on this particle? (b) At what acceleration will this charge q move at this location? (Show all work!)
x
Q 1 > 0
x 1
Q 2 > 0
x 2
q < 0
xq
F1q F2q
Solution:
Part (a): Since Q 1 > 0 and q < 0, the force (F 1 q) between Q 1 > 0 and q < 0 points to the left (as shown in the figure above) due to the fact that these charges will attract each other. Since Q 2 > 0 and q < 0, the force (F 2 q) between Q 2 > 0 and q < 0 points to the right, once again due to the fact that these charges will attract each other. As stated in the problem, the sum of these two forces is +5. 56 × 10 −^4 N (Ftot). Using this fact in conjunction with Coulomb’s Law, we can do a little algebra to solve for q:
Ftot = F 2 q − F 1 q
= ke
|Q 2 | |q| r^22 q
− ke
|Q 1 | |q| r^21 q
= ke |q|
[ |Q 2 | r 22 q
|Q 1 |
r^21 q
] ,
|q| = Ftot ke
[ |Q 2 | r^22 q
|Q 1 |
r 12 q
]− 1 .
r 1 q = xq − x 1 = 2.89 m − (− 4 .56 m) = 7.45 m r 2 q = (x 2 − xq) = 8.05 m − 2 .89 m = 5.16 m |Q 1 | r^21 q
6. 88 × 10 −^5 C
(7.45 m)^2
6. 88 × 10 −^5 C
55 .5025 m^2
= 1. 24 × 10 −^6 C/m^2.
|Q 2 | r^22 q
2. 33 × 10 −^4 C
(5.16 m)^2
2. 33 × 10 −^4 C
26 .6256 m^2
= 8. 75 × 10 −^6 C/m^2.
|q| =
5. 56 × 10 −^4 N
- 99 × 109 N · m^2 /C^2
[
- 75 × 10 −^6 C/m^2 − 1. 24 × 10 −^6 C/m^2
]− 1
= 8. 24 × 10 −^9 C.
Since q has to be negative as stated in the problem, q = − 8. 24 × 10 −^9 C.
Part (b): Here we just use Newton’s Second Law of Motion: F = m a, and solve for a the acceleration (remembering to change the mass from cgs units to SI units):
a =
Ftot m
5. 56 × 10 −^4 N
- 65 × 10 −^4 kg
= 0 .727 m/s^2.
- A proton is released from rest at x = − 4 .56 m in a uniform electric field of +6. 68 × 106 N/C ˆx. (a) Calculate the change in potential energy when the proton moves along the x-axis to the x = +7.58 m position. (b) What will be the velocity of the proton at this position of +7.58 m? (c) What is the acceleration of this proton as it moves through the electric field? (Show all work!)
x 0 x o xf
E = 6.68 x 10^6 N/C
q = +e mp v
Solution:
Part (a): The figure to the right diagrams the problem, where our particle (a proton), has mass mp = 1. 672 × 10 −^27 kg, charge q = +e = 1. 602 × 10 −^19 C, and travels along the x-axis from x◦ = -4.56 m (where it starts from rest) to xf = 7.58 m for a total distance of d = xf − x◦ = 7.58 m - (-4.56 m) = 12.14 m. This proton travels in an electric field E of strength 6. 68 × 106 N/C which runs parallel to the x axis. The change in potential energy can now be determined with:
∆PE = −q E d = −(1. 602 × 10 −^19 C) (6. 68 × 106 N/C) (12.14 m) = − 1. 30 × 10 −^11 J ,
where the negative sign results from the fact that q > 0 (i.e., a positive value).
Part (b): To determine the final velocity, the simplest way is to just use the conservation of energy. Here, we do not need to include the gravitational potential energy since the mass just moves in the horizontal direction, hence PEg i = PEg f. Since we are starting from rest, the initial kinetic energy is equal to 0.
KE (^) i + PE (^) i = KE (^) f + PE (^) f PE (^) i − PE (^) f = KE (^) f − KE (^) i −∆PE =
m vf^2 − 0
v^2 f =
−2(∆PE)
mp
vf =
√ −2(− 1. 30 × 10 −^11 J)
- 672 × 10 −^27 kg
vf = 1. 25 × 108 m/s.
- In the figure below, let Q 1 = +6.84 μC, Q 2 = -16.2 μC, and Q 3 = +9.20 μC be rigidly fixed and separated by r 12 = r 13 = r 23 = 1.34 m. Calculate the electric potential at point P which is half-way between Q 1 and Q 2. (Show all work!)
Q 1 Q 2
Q 3
P
r 12
r 13 r 23
x
Solution: y
First we need to calculate the distances that point P is from each charge. We will define r 1 as the distance from Q 1 to P, r 2 as the distance from Q 2 to P, and r 3 as the distance from Q 3 to P — note that r 3 is perpendicular to r 12 :
r 1 = 12 r 12 = 12 (1.34 m) = 0.670 m r 2 = 12 r 12 = 12 (1.34 m) = 0.670 m
r 3 =
√ r^223 − r^22 = 1.16 m
Now we use the superposition principle (Eq. II-8) to calculate the total potential at point P:
V = ke
∑^3 i=
( (^) q i ri
)
= ke
( Q 1 r 1
Q 2
r 2
Q 3
r 3
)
(
- 99 × 109 N m^2 /C^2
) [^6. 84 × 10 −^6 C
0 .670 m
− 16. 2 × 10 −^6 C
0 .670 m
9. 20 × 10 −^6 C
1 .16 m
]
(
- 99 × 109 N m^2 /C^2
) [ − 6. 04 × 10 −^6 C/m
]
= − 5. 43 × 104 V.
- A fully charged capacitor stores 6. 89 × 106 eV of energy while connected to a 1.50 V battery. (a) If this capacitor has circular plates of radius 1.34 cm, what is the area on one of the plates? (b) What is the capacitance of this capacitor? (c) What must be the separation between these two plates (in mm)? (d) How much charge is on these plates when fully charged? (Show all work including units!)
Solution:
Part (a): Since there are no other components in this circuit, the potential across the plates of the capacitor is the same as the potential delivered by the battery: ∆V = 1.50 V. We now need to convert the energy of this capacitor from eVs to SI units:
U = 6. 89 × 106 eV · 1. 602 × 10 −^19 J/eV = 1. 10 × 10 −^12 J.
To calculate the area, we first must convert the radius from cgs to SI units, r = 1.34 cm =
- 34 × 10 −^2 m. For a circular area, we get
A = π r^2 = π (1. 34 × 10 −^2 m)^2 = 5. 64 × 10 −^4 m^2.
Part (b): We determine the capacitance using the capacitor energy equation (Eq. II-28):
U =
C (∆V )^2
C =
2 U
(∆V )^2
2 (1. 10 × 10 −^12 J)
(1.50 V)^2
= 9. 81 × 10 −^13 F = 0.981 pF.
Part (c): For the separation use Eq. (II-14):
C = ◦
A
d
d = ◦
A
C
(
- 85 × 10 −^12 C^2 /N m^2
) (^5). 64 × 10 − (^4) m 2
- 81 × 10 −^13 F = 5. 09 × 10 −^3 m = 5 .09 mm.
Part (d): We can now use either Eq. (II-12) or Eq. (II-28) to calculate the full charge, here we will use Eq. (II-12) first, then Eq. (II-28):
C =
Q
∆V
Q = C ∆V = (9. 81 × 10 −^13 F) (1.50 V) = 1. 47 × 10 −^12 C.
U =
Q ∆V
Q =
2 U
∆V
2 (1. 10 × 10 −^12 J)
1 .50 V
= 1. 47 × 10 −^12 C.
