Coulomb\'s law for the magnitude of the force F between two particles with charg

Question

Coulomb's law for the magnitude of the force Fbetween two particles with chargesQ and Q?separated by a distance d is

|F|=K|QQ?|d2,

where K=14??0, and ?0=8.854×10?12C2/(N?m2) is the permittivity of free space.

Part A

What is the net force exerted by these two charges on a third charge q3 = 52.5 nC placed between q1and q2 at x3 = -1.080 m ?

Your answer may be positive or negative, depending on the direction of the force.

Consider two point charges located on the x axis: one charge, q1 = -12.0 nC , is located atx1 = -1.735 m ; the second charge,q2 = 34.5 nC , is at the origin

my last try

Forces in a Three-Charge System Part A Part A What is the net force exerted by these two charges on a third charge g-52.5 nC placed between Coulomb's law for the magnitude of the force F between two particles with charges and separated by a distance d is and at =-1.080 ? Your answer may be positive or negative, depending on the direction of the force Express your answer numerically in newtons to three significant fiqures where K = , and 40 E0-8854 × 10 12 C2/(N-m2) is the permittivity of free space Force on q3 Consider two point charges located on the x axis one charge, q1 =-12.0 nC , is located at T1 1.135 m, the second charge q2-34.5 nC , is at the origin (z-0.0000) Submit Hints My Answers Give Up Review Part ide Feedbac Continue

Explanation / Answer

r2 = distance of charge q2 from q3 = 0 - (- 1.08) = 1.08 m

r1 = distance of charge q1 from q3 = - 1,08 - (- 1.735) = 0.66 m

Force by q1 on q3

F1 = k q1 q3 / r12 = (9 x 109) (12 x 10-9) (52.5 x 10-9) / (0.66)2 = 0.000013 N   towards left

Force by q2 on q3

F2 = k q2 q3 / r22 = (9 x 109) (34.5 x 10-9) (52.5 x 10-9) / (1.08)2 = 0.000013975 N   towards right

Net force = F2 - F1 = 0.000013975 - 0.000013 = 9.75 x 10-7

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