Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig
ht as positive and forces pointing to the left as negative. Rank in increasing order by placing the most negative on the left and the most positive on the right. To rank items as equivalent, overlap them.
1 answer:
I have attached the image showing the 6 combination of charges
Answer:
Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc
Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc
Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc
Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc
Explanation:
The electrostatic force between 2 charges is; F = k•q1•q2/r²
Where r is the distance between the 2 charges q1 and q2 and k is coulombs constant.
Thus;
F ∝ q1•q2/r²
The net force on the charge q1 is the sum of the forces on charge q2 and q3.
If the two charges are opposite, the force is an attractive force while if the two charges are similar, the force is a repulsive force.
Thus, if the combination has same type of charges, i.e;(+1,+1,+1) or (-1,-1,-1), the force will be very small.
So, now the order of forces from smallest to largest is;
Smallest are: q1 = +1 nc, q2 = +1 nc, q3 = +1 nc and q1 = -1 nc, q2 = -1 nc, q3 = -1nc
Third Largest: q1 = +1 nc, q2 = +1nc, q3 = -1 nc
Second Largest: q1 = +1 nc, q2 = -1 nc, q3 = +1 nc
Largest are: q1 = +1 nc, q2 = -1 nc, q3 = -1 nc and q1 = -1 nc, q2 = +1 nc, q3 = +1 nc
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Answer:
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω
Explanation:
Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.
When the current increases by 0.450 A wen only R1 is in the circuit, the current is
9/R1 + R2 + 0.450 A = 9/R1 (1)
When the current increases by 0.225 A when only R2 is in the circuit, the current is
9/R1 + R2 + 0.225 A = 9/R2 (2)
equation (1) - (2) equals
9(1/R1 - 1/R2) = 0.450 A - 0.225
9(1/R1 - 1/R2) = 0.125
(1/R1 - 1/R2) = 0.125 A/9 = 0.0138
1/R1 = 0.0138 + 1/R2
R1 = R2/(1 + 0.0138R2) (3)
From (1)
9/R1 - 9/R1 + R2 = 0.450 A
9R2/[R1(R1 + R2)] = 0.450 A
R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5
R2/[R1(R1 + R2)] = 0.5 (4)
From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,
(1 + 0.0138R2) = 0.5(R1 + R2)
0.5R1 + 0.5R2 = 1 + 0.0138R2
0.5R1 = 1 + 0.0138R2 - 0.5R2
0.5R1 = 1 - 0.4862R2 (5)
Substituting (3) into (5) we have
0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2
R2 = (1 + 0.0138R2)(1 - 0.4862R2)
R2 = 1 - 0.4724R2 - 0.0067R2²
Collecting like terms, we have
0.0067R2² + 0.4724R2 + R2 - 1 = 0
0.0067R2² + 1.4724R2 - 1 = 0
Using the quadratic formula,
We choose the positive answer.
So R2 = 0.340 Ω
From (5)
R1 = 0.5 - 0.9931R2
= 0.5 - 0.9931 × 0.340
= 0.5 - 0.338
= 0.162 Ω
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω