Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1457483 • Letter: C
Question
Coulomb's law for the magnitude of the force Fbetween two particles with charges Q and Qseparated by a distance d is
|F|=K|QQ|d2,
where K=140, and 0=8.854×1012C2/(Nm2) is the permittivity of free space.
Consider two point charges located on the x axis: one charge, q1 = -12.5 nC , is located atx1 = -1.675 m ; the second charge,q2 = 34.0 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 49.0 nC placed between q1and q2 at x3 = -1.135 m ?
Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
the distance between charge 3 and charge 1 is,
r13 = x3 - x1 = -1.135 m + 1.675 m = 0.54 m
the force F31 is,
F31 = kq1q3/r132 = 9x109*12.5x10-9*49x10-9 / 0.54*0.54 = 1.89x10-5 N, toward charge q1
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The force F32 is,
F32 = kq2q3/r232 = 9x109*34x10-9*49x10-9/ 1.135*1.135 = 1.16x10-5 N , toward charge q1.
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Net force is,
Fnet = F1 + F2 = 1.89x10-5 N + 1.16x10-5 N = 3.05x10-5 N, to the left
vector form:
Fnet = -3.05x10-5 N i^