Two point charges q 1 and q 2 are held in place 4.50 c m apart. Another point ch
ID: 2269658 • Letter: T
Question
Two point charges q1 and q2 are held in place 4.50 cm apart. Another point charge -1.80?C of mass 5.00g is initially located 3.00 cm from each of these charges (the figure) and released from rest. You observe that the initial acceleration of -1.80?C is 324m/s2 upward, parallel to the line connecting the two point charges.
Find q1 and q2.
Two point charges q1 and q2 are held in place 4.50 cm apart. Another point charge -1.80?C of mass 5.00g is initially located 3.00 cm from each of these charges (the figure) and released from rest. You observe that the initial acceleration of -1.80?C is 324m/s2 upward, parallel to the line connecting the two point charges. Find q1 and q2.Explanation / Answer
In order for Q to move upward it must be attracted to q1 and repelled by q2. q1 and q2 must also have the same charge, but opposite signs as just mentioned, so that the perpendicular force will vanish.
Just a reminder, Coulomb's Law, will be needed:
F = kQ1*Q2/r^2
k = 9.0x10^9 Nm^2/C^2
Upward force on Q is (first convert mass from g to kg):
F = ma = (0.005)(324) = 1.62 N
If Fv is the vertical component of force from q1 then the force from q2 is also Fv. So the total force is 2Fv and it must equal F just calculated.
2Fv = 1.62
Fv = 0.81 N
This is related to the total force between q1 and q by:
cos(A) = Fv/(total force)
A = the angle between an upward vector at Q to the line from Q to q1.
cos(A) = 2.25/3 = 0.75
total force = Fv/cos(A) = 0.81/0.75 = 1.08 N
Using Coulombs law (stated at the start) and first converting charge from microC to C and distance in cm to m:
1.08 = (9.0x10^9)[1.8x10^(-6)][q1]/(.03)^2
q1 = 0.6x 10^(-7) C
We can the write:
q1 = +0.06 microC
q2 = -0.06 microC
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