No Figure was given for these two problems. 3 charges, 5 µC each, are located on

Question

No Figure was given for these two problems.

3 charges, 5 µC each, are located on three vertices A, B, C of an equilateral triangle with sides 1 cm each. Another charge q is located at the mid point D of the side BC. Calculate q in micro Coulomb so that net force on the charge at A due to the charges at B, C and D is zero.

In a right angle traingle ABC, angle ABC is 90 Degree, AB = 2 m, and angle ACB is 41.81 Degree. A point charge of 5*24 nC is placed at point C, point charge 4* 24 nC is placed at point A and point charge 1 C is placed in point B. Calculate the force on charge at B due to others two.

Explanation / Answer

1) Length of AD, rAD = (1 * 10-2) * sin60o = 8.66 * 10-3 m

Net force on charge at A, Fnet = kQ[(2Qcos30o/rAB2) + q/rAD2] = 0

=> 2 * (5 * 10-6) * cos30o / (1 * 10-2)2 = -q / (8.66 * 10-3)2

=> q = -6.5 µC

2) BC = AB/tan41.81o = 2.24 m

Net force on B, Fnet = FBCi - FABj = kQB[(QC/BC2)i - (QA/AB2)j]

=> Fnet = (9 * 109) * 1 * [(5.24 / 2.242)i - (4.24 / 22)j] * 10-9 = 9.40i - 9.54j

|Fnet| = [(9.40)2 + (-9.542)]1/2 = 13.4 N

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