A charge of 20.2 micro coulombs is held fixed at the origin. If a -5.25 micro co

Question

A charge of 20.2 micro coulombs is held fixed at the origin. If a -5.25 micro coulomb charge with a mass of 3.2 grams is released from rest at the position (0.925m, 1.17m), what is its speed when it is halfway to the origin?

My working so far:
For the second charge:
Ui = Uf + Kf
Kf = -(Uf - Ui)
= -(Potential Difference) X -5.25mC

But I'm lost at calculating potential difference between these 2 points, because I can't use kq/r (no r as a basis for comparison). Can somebody guide me with regards to this point?

Explanation / Answer

20.2uC (0,0) -5.25uC and 3.2gm

F=ma
Q1 Q2 x9x10^9/(.9252 + 1.172) = 3.2x10^-3 a   

a= 199.98m/s2

v2 = 2as

   = 2x199.98 x (.9252 + 1.172)/2

v= 17.27m/s

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