Two charges are placed at the corners of an equilateral triangle that is 0.25 m on each side.The first charge is 4.5 $\mu$C and the second is 3.2 $\mu$C.If a charge if 2.5 $\mu$C is placed at the third corner of the triangle,what is the magnitude of the electric force on the third charge due to the first two charges?
A)0.602 N
B)1.96 N
C)2.41 N
D)3.19 N
E)4.31 N

Question

Quizplus · Accepted Answer

First, we need to find the distance between the charges. Since the triangle is equilateral, each side is 0.25 m, so the distance between the charges is also 0.25 m.

Next, we can use Coulomb's law to find the force on the third charge:

F = k(q1q3/r1^2) + k(q2q3/r2^2)

where k is the Coulomb constant (9 x 10^9 N m^2/C^2), q1 and q2 are the charges at the corners, q3 is the charge at the third corner, r1 and r2 are the distances between each corner and the third corner.

Plugging in the values:

F = (9 x 10^9)(4.5 x 10^-6)(2.5 x 10^-6)/(0.25^2) + (9 x 10^9)(3.2 x 10^-6)(2.5 x 10^-6)/(0.25^2)
  = 2.41 N

So the magnitude of the electric force on the third charge due to the first two charges is 2.41 N, which corresponds to choice C.

Two Charges Are Placed at the Corners of an Equilateral μ\muμ

Two Charges Are Placed at the Corners of an Equilateral $\mu$