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Physics help?
In a vacuum, two particles have charges of q1 and q2, where q1 = +3.4C. They are separated by a distance of 0.28 m, and particle 1 experiences an attractive force of 4.9 N. What is the value of q2, with its sign?
2 Answers
- oldschoolLv 74 years ago
F = kq1q2/r² = 4.9N ------> q2 = 4.9*r²/kq1 = 4.9*0.28²/(9e9*3.4) = 12.6e-12 = -12.6pC and must be negative because opposite polarity charges attract.
- 4 years ago
Coulomb's Law relating force (F) between two charges (q1, q2) separated by a distance (r):
F = κ.q1.q2 / r²
where Coulomb constant κ = 8.99 x 10^9 N.m² /C²
re-arrange:
q2 = | F.r² / κ.q1 |
q2 = | 4.9 * 0.28² / ( 8.99 x 10^9 * 3.4) |
q2 = | 1.26 x 10^-11 C |
The force is attractive, so q1 and q2 have opposite signs
q2 ≈ -1.3 x 10^-11 c
