A tiny sphere of mass 740 mu g and charge 280 nC is initiall

A tiny sphere of mass 7.40 mu g and charge -2.80 nC is initially at a distance of 1.44 pm from a fixed charge of +7.73 nC. If the 7.40-mu g sphere is released from rest, find its kinetic energy when it is 0.500 pm from the fixed charge. If the 7.40-mu g sphere is released from rest, find its speed when it is 0.500 mu m from the fixed charge.

Solution

Given: mass of sphere m = 7.4 g ; charge on sphere q = -2.8 nC ; fixed charge Q = 7.73 nC ;

initial distance d_i = 1.44 m and final distance = 0.5 m

Now applying energy conservation:

increase in the kinetic energy of the sphere = decrease in it\'s potential energy i,e

(a) KE = kQq/d_i - kQq/d_f = -9*10⁹*2.8*7.73*10^-18*(1/4 - 1/0.5)*10⁶ = 0.34 J

(b) 1/2*m*v² = KE or v = (2*KE/m)^0.5 = (2*0.34/7.4*10^-9)^0.5 = 9586 m/s