Point charges of 4.0 $\times$ 10^-8 C and -2.0 $\times$ 10^-8 C are placed 12 cm apart.A third point charge of 3.0 $\times$ 10^-8 C halfway between the first two point charges experiences a force of magnitude A)4.5 $\times$ 10^-3 N B)2.0 $\times$ 10^-3 N C)1.5 $\times$ 10^-3 N D)zero E)5.0 $\times$ 10^-3 N

Question

Quizplus · Accepted Answer

Using Coulomb's law, we can find the force experienced by the third charge:
F = k(q1q3)/(r1^2) + k(q2q3)/(r2^2)
where k is Coulomb's constant, q1 and q2 are the charges of the first two point charges, q3 is the charge of the third point charge, r1 is the distance between the first point charge and the third point charge, and r2 is the distance between the second point charge and the third point charge.

Plugging in the given values, we get:

F = (9x10^9)(4x10^-8)(3x10^-8)/(0.06)^2 + (9x10^9)(-2x10^-8)(3x10^-8)/(0.06)^2
F = 4.5x10^-3 N

Therefore, the magnitude of the force experienced by the third charge is 4.5x10^-3 N, which is closest to choice A (4.5x10^-3 N).

Point Charges of 4 ×\times× 10-8 C and -2 ×\times× 10-8 C Are Placed 12 Cm Apart

Point Charges of 4 $\times$ 10^-8 C and -2 $\times$ 10^-8 C Are Placed 12 Cm Apart