Two particles, each having a mass of 3.0 m g and having equal but opposite charg

Q: Two particles, each having a mass of 3.0 m g and having equal but opposite charg

apply KE = change in PE 0.5 mv^2 + 0.5 mv^2 = Kq1q2/r since both masses are same and V is same we get mv^2 = kq1q2/r r = Kq1q2/mv^2 r = (9e9 * 6 e-9 * 6e -9 )/(3 e-6 * 5*5) r = 4.32 mm----------

ID: 1407384 • Letter: T

Question

Two particles, each having a mass of 3.0 mg and having equal but opposite charges of magnitude of 6.0 nC, are released simultaneously from rest when they are a very large distance apart. What distance separates the two at the instant when each has a speed of 5.0 m/s?

The answer is 4.32 nm but can you please explain in detail, why?

Explanation / Answer

apply KE = change in PE

0.5 mv^2 + 0.5 mv^2 = Kq1q2/r

since both masses are same and V is same

we get mv^2 = kq1q2/r

r = Kq1q2/mv^2

r = (9e9 * 6 e-9 * 6e -9 )/(3 e-6 * 5*5)

r = 4.32 mm----------<<<<<<<<<<<<<<Answer

Note: Do check ur answer in terms of nm

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Two particles, each having a mass of 3.0 m g and having equal but opposite charg

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