A 31.91-mC charge is placed 34.65 cm to the left of a 77.73-mC charge, as shown
ID: 1524379 • Letter: A
Question
A 31.91-mC charge is placed 34.65 cm to the left of a 77.73-mC charge, as shown m the figure, and both charges are held stationary A part tide with a charge of -7.051 mu C and a mass of 23.81 g (depicted as a blue sphere) is placed at rest at a distance 31.19 cm above the right-most charge. If the particle were to be released from rest, it would follow some complicated path around the two stationary charges Calculating the exact path of the particle would be a challenging problem, but even without performing such a calculation it is possible to make some definite predictions about the future motion of the particle. If the path of the particle were to pass through the gray point labeled A. what would be its speed v_A at that point?Explanation / Answer
using the Formula
work done in dispalcing the charge q = -7.051uC from the given position to the A is W = U1-U2
U1 is the initial potential energy of the system
U2 is the final potential energy of the system
U1 = k*[(q1*q2/r1)+(q1*q3/r2)+(q2*q3/r3)]
U1 = k*[((77.73*31.91*10^-6)/(0.3465)) - ((77.73*7.051*10^-9)/(0.3119)) - ((31.91*7.051*10^-9)/(sqrt(0.3465^2+0.3119^2))) ]
U1 = 9*10^9*7.15*10^-3 J = 6.435*10^7 J
U2 = k*[(q1*q2/r1)+(q1*q3/r2)+(q2*q3/r3)]
U2 = k*[((77.73*31.91*10^-6)/(0.3465)) - ((77.73*7.051*10^-9)/(0.2425)) - ((31.91*7.051*10^-9)/(0.1040))]
U2 = 6.4385*10^7 J
W = U1 - U2
W = (6.435*10^7) - (6.4385*10^7)
W = -35000 J
W = 3.5*10^4 J
and also using work energy theorem
W = K2-K1
W = 0.5*m*v^2
3.5*10^4 = 0.5*23.81*10^-3*v^2
v = 1.72*10^3 m/sec
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