2 equal charges, 174 micro coulombs each, experiences an attraction force of 174

Question

2 equal charges, 174 micro coulombs each, experiences an attraction force of 174 newton. Calculate the distance between the charges in mm. what is the total charge of all protons in 3 gram of water (H_2O)? 3 charges, 7 muC 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 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 triangle ABC, angle ABC is 90 degree, AB = 2m, and angle ACB is 41.81 Degree. A point charge of 5Degree9 nC is placed at point C, point charge 4Degree 9nC 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

15) F = kq2/r2

=> r = q(k/F)1/2 = (174 * 10-6) * (9 * 109 / 174)1/2 = 1.25 * 106 mm

16) Number of moles in 3 g of water, n = 3/18 = 1/6 moles

There are (8 + 1 + 1) = 10 protons in one water molecule.

Total charge of all protons, Q = (1/6) * (6.023 * 1023) * 10 * (1.602 * 10-19) = 1.6 * 105 C

17) 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 * (7 * 10-6) * cos30o / (1 * 10-2)2 = -q / (8.66 * 10-3)2

=> q = -9.1 µC

18) 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 * [(5.9 / 2.242)i - (4.9 / 22)j] * 10-9 = 10.6i - 11.0j

|Fnet| = [(10.6)2 + (-11.02)]1/2 = 15.3 N

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.