A 15.41-mC charge is placed 33.15 cm to the left of a 53.73-mC charge, as shown

Question

A 15.41-mC charge is placed 33.15 cm to the left of a 53.73-mC charge, as shown in the figure, and both charges are held stationary. A particle with a charge of -6.051 /l/C and a mass of 39.81 g (depicted as a blue sphere) is placed at rest at a distance 29.84 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

let

Q1 = 15.41 mC

Q2 = 53.73 mC

q = -6.051 micro C

Potential at initial position,

Vi = k*Q1/d1 + k*Q2/d2

= 9*10^9*15.41*10^-3/sqrt(0.3315^2 + 0.2984^2) + 9*10^9*53.73*10^-3/0.2984

= 1.93*10^9 volts

Potential at final position,

Vf = k*Q1/d1 + k*Q2/d2

= 9*10^9*15.41*10^-3/0.09945 + 9*10^9*53.73*10^-3/(0.3315 - 0.09945)

= 3.478*10^9 volts

Apply conservation of energy

ki + Ui = Kf + Vf

0 + Ui = Kf + Uf

Kf = Ui - Uf

= q*Vi - q*Vf

Kf = q*(Vi - Vf)

0.5*m*v^2 = q*(Vi - Vf)

v = sqrt(2*q*(Vi - Vf)/m)

= sqrt(2*(-6.051*10^-6)*(1.93 - 3.478)*10^9/(39.81*10^-6))

= 2.169*10^4 m/s

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