Two equally charged particles, held 8.5 mm apart, are released from rest. The in

Question

Two equally charged particles, held 8.5 mm apart, are released from rest.  The initial magnitude of the acceleration of the first particle is observed to be 3.818*10^-15 m/s^2 and the second to be 2.846*10^-15 m/s^2. if the mass of the first particle is 41 micrometers (ug?),

1. what is the mass of the second?

2. what is the magnitude of the common charge?

3. what is the force on each particle when the distance between them is halved?

I can't seem to get the correct conversions, thus wrong exponent for the answer.  Plz help! Explain as well! Thanx

Explanation / Answer

f = kq1q2/r^2

f/m1 = 3.818 *10^-15 f = 3.818*10^-15 *41 = 156.538 *10^-15

f/m2 = 2.4846*10^-15 m2 = 3.818*10^-15 *41/2.4846*10^-15 = 63.0033

2) q=sqrt (fr^2/k) , sqrt (156.538 *10^-15 *72.25 *10^-6 /9 *10^9 ), =sqrt (1256.65 *10^-30) = 35.44*10^-15 C

3)

if the distance is halved force becomes 4 times

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