Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1531347 • Letter: C
Question
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q separated 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 = -17.0 nC , is located at x1 = -1.745 m ; the second charge, q2 = 39.0 nC , is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q3 = 50.5 nC placed between q1 and q2 at x3 = -1.090 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
The answer is NOT -3.08x10^-6
Explanation / Answer
Let F be the force exerted by q and q; let F be the force exerted by q and q; let r be the distance between q and q: 0.655 m; let r be the distance between q and q: 1.090 m.
F = kFqq/r²
= (9 × 10 N·m²/C²)(-17.0 × 10 C)(50.5 × 10 C) / (0.655 m)²
= -1.80 × 10 N.
F = kFqq/r²
= (9 × 10 N·m²/C²)(39.0 × 10 C)(50.5 × 10 C) / (1.090 m)²
= +1.49 × 10 N.
The net force is F + F = -1.80 × 10 N + 1.49 × 10 N= 0.31x10^-5 N ( in the negative x-direction)