Two particles, with charges of q_1 = 60.0 nC and q_2 = -60.0 nC, are placed at t

Question

Two particles, with charges of q_1 = 60.0 nC and q_2 = -60.0 nC, are placed at the points with coordinates (0, 4.00 cm) and (0, -4.00 cm) as shown in the figure below. A particle with charge q_3 = 30.0 nC is located at the origin. Find the electric potential energy of the configuration of the three fixed charges. A fourth particle, with a mass of 2.04 times 10^-13 kg and a charge of q_4 = 120.0 nC, is released from rest at the point (3.00 cm, 0). Find its speed after it has moved freely to a very large distance away. m/s

Explanation / Answer

part a:

electric potential energy =k*((q1*q2/d12)+(q2*q3/d23)+(q3*q1/d13))

where k=coloumb's constant=9*10^9

q1=60*10^(-9) C

q2=-60*10^(-9) C

q3=30*10^(-9) C

d12=distance between q1 and q2=8 cm=0.08 m

d23=distance between q2 and q3=4 cm=0.04 m

d13=distance between q1 and q3=0.04 m

hence electrical potential energy of the configuration

=9*10^9*((60*10^(-9)*(-60)*10^(-9)/0.08)+(30*10^(-9)*(-60)*10^(-9)/0.04)+(60*10^(-9)*(30)*10^(-9)/0.04))

=-4.05*10^(-4) J

part b:

distance of q4 from q1=d14=sqrt(4^2+3^2)=5 cm=0.05 m

distance of q4 from q2=d24=sqrt(4^2+3^2)=5 cm =0.05 m

distance of q4 from q3=d34=3 cm=0.03 m

electric potential at the location of q4=k*((q1/d14)+(q2/d24)+(q3/d34))

=9*10^9*((60*10^(-9)/0.05)+((-60)*10^(-9)/0.05)+(30*10^(-9)/0.03))=9000 volts

so when the particle has moved to large distance away,

electrical potential energy will be converted to kinetic energy

hence charge*potential =0.5*mass*speed^2

==>speed=sqrt(2*charge*potential/mass)

=sqrt(2*120*10^(-9)*9000/(2.04*10^(-13)))

=1.029*10^5 m/s

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