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Physics Help?
In a vacuum, two particles have charges of q1 and q2, where q1 = +3.4UC. 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?
1 Answer
- Anonymous4 years ago
Each particle is attracted to the other with a force of 4.9N.
F = k.q1.q2/d² Attraction is treated as negative.
Putting the numbers in gives:
-4.9 = 9x10⁹ x 3.4x10⁻⁶ x q2 / 0.28²
q2 = - 4.9 x 0.28² / (9x10⁹ x 3.4x10⁻⁶)
. . .= -1.26 x 10⁻⁵ C
. . .= -12.6 x 10⁻⁶ C
(-12.6μC or -13μC to 2 significant figures)