Particles of charge Q 1 = +61 C, Q 2 = +55 C, and Q 3 = -80 C are placed in a li
ID: 1325021 • Letter: P
Question
Particles of charge Q1 = +61 C, Q2 = +55 C, and Q3 = -80 C are placed in a line (Fig. 16-49). The center one is 0.35 m from each of the others.
Calculate the net force on each charge due to the other two.
Force on Q1_______ N
Force on Q2_______N
Force on Q3_______N
Force on Q1_______ N
Force on Q2_______N
Force on Q3_______N
Particles of charge Q1 = +61 C, Q2 = +55 C, and Q3 = - 80 C are placed in a line (Fig. 16 - 49). The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two. Force on Q1_______ N Force on Q2_______N Force on Q3_______NExplanation / Answer
On Q1:
Force due to Q2:
Note that
F = k q1 q2 / r^2
where
k = 8.99E+09 N m^2/C^2
F = the force of interaction
q1 = the first charge = 6.10E-05 C
q2 = the second charge = 5.50E-05 C
r = the distance between q1 and q2 = 0.35 m
Thus,
F = -246.2159184 N, to the left
Force due to Q3:
Note that
F = k q1 q2 / r^2
where
k = 8.99E+09 N m^2/C^2
F = the force of interaction
q1 = the first charge = 6.10E-05 C
q2 = the second charge = -8.00E-05 C
r = the distance between q1 and q2 = 0.7 m
Thus,
F = +89.53306122 N, to the right
Thus, the net force is the sum of these two forces,
F(Q1) = 157 N [to the left] [ANSWER]
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On Q2:
Force due to Q1:
Note that
F = k q1 q2 / r^2
where
k = 8.99E+09 N m^2/C^2
F = the force of interaction
q1 = the first charge = 6.10E-05 C
q2 = the second charge = 5.50E-05 C
r = the distance between q1 and q2 = 0.35 m
Thus,
F = 246.2159184 N, to the right
Force due to Q3:
Note that
F = k q1 q2 / r^2
where
k = 8.99E+09 N m^2/C^2
F = the force of interaction
q1 = the first charge = 5.50E-05 C
q2 = the second charge = -8.00E-05 C
r = the distance between q1 and q2 = 0.35 m
Thus,
F = 322.9061224 N
Thus, the sum of these two forces are
F(Q2) = 569.1 N , to the right [ANSWER]
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On Q3:
Force due to Q1:
Note that
F = k q1 q2 / r^2
where
k = 8.99E+09 N m^2/C^2
F = the force of interaction
q1 = the first charge = 6.10E-05 C
q2 = the second charge = -8.00E-05 C
r = the distance between q1 and q2 = 0.7 m
Thus,
F = -89.53306122 N, to the left
Force due to Q2:
Note that
F = k q1 q2 / r^2
where
k = 8.99E+09 N m^2/C^2
F = the force of interaction
q1 = the first charge = 5.50E-05 C
q2 = the second charge = -8.00E-05 C
r = the distance between q1 and q2 = 0.35 m
Thus,
F = -322.9061224 N, to the left
Thus,
F(Q3) = -412.4 N, to the left [ANSWER]
**************
SUMMARY:
Force on Q1: 157 N, to the left
Force on Q2: 569.1 N, to the right
Force on Q3: 412 N, to the left