The air in living room has a mass of 72 kg and a specific heat of 1010 J/(kg°C). What is the approximate change in thermal energ
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Answer:
1) U = -4.436 10-7 J , 2) r13 = 0.6937 m
Explanation:
The electric potential energy is given by the equation
U = k Σ
/
1. Let's apply this equation to our case
U = k (q1q3 / r13 + q1 q2 / r12 + q2 q3 / r23)
Let's reduce all magnitudes to the SI system
q1 = 4.10 nC = 4.10 10-9 C
q2 = -2.90 nC = -2.90 10-09 C
q3 = 2.00 nC = 2.00 10-9 C
Let's look for the distances
r13 = 11-0 = 11 cm = 11 10-2 m
r12 = 20-0 = 20 cm = 20 10-2 m
r23 = 20 -11 = 9 cm = 9.0 10-2 m
Let's calculate the electric potential
U = 8.99 109 [(4.10 2.00) 10-18 / 11 10-2 + (4.10 (-2.90)) 10-18 / 20 10-2 + (-2.90 2.00) 10-18 / 9.0 10-2]
U = 8.99 [0.7455 - 0.5945 - 0.6444) 10-7
U = -4.436 10-7 J
2. ask to find the position of q3 for the energy to be zero
U = 0 = k (q1q3 / r13 + q1 q2 / r12 + q2 q3 / r23)
In this case the distance between 1 and 2 is fixed, since the load that is placed is 3
q1 q2 / r12 = - q1 q3 / r13 - q2 q3 / r23
-11.89 10-18 / 20 10-2 = -8.20 10-18 / r13 + 5.810-18 / r23
-0.5945 102 = -8.20 / r13 + 5.810 / r23
Let's relate the distances, the maximum separation esr12 which is 20 10-2 m
r12 = r13 + r23
r13 = r12 -r23
r13 = 2010-2 - r23
we replace
-0.5945 102 = -8.20 / (20 10-2 - r23) + 5.810 / r23
Let's solve this equation
-0.5945 102 (2010-2 - r23) = -8.20 r23 + 5.810 (2010-2 - r23)
- 11.89 + 0.5945 r23 = -8.20 r23 + 1,162 - 5,810 r23
-11.89 - 1,162 = - 8.20 r23 -5,810 r23 - 0.5945 r23
-13.052 = r23 (-14.6045)
r23 = 13.052 / 14.6045
r23 = 0.8937 m
The distance from the origin is
r13 = -r12 + r23
r13 = -20 10-2 + 0.8937
r13 = 0.6937 m