The answers are not 42.89 x10^3, 1.84x10^6, 1.35x10^3. The answer has to be in m

Question

The answers are not 42.89 x10^3, 1.84x10^6, 1.35x10^3. The answer has to be in m/s.

The answers are not 42.89 x10^3, 1.84x10^6, 1.35x10^3. The answer has to be in m/s. Your answer is partially correct. Try again. Two identical point charges (q = +9.00 x 10^-6 C) are fixed at opposite corners of a square whose sides have a length of 0.550 m. A test charge (q0 = -5.40 x 10^-8 C), with a mass of 1.10 x 10^-8 kg, is released from rest at one of the corners of the square. Determine the speed of the test charge when it reaches the center of the square. Number 1.35 Units m/s the tolerance is + / -2% Click if you would like to Show Work for this question: Open Show Work

Explanation / Answer

side of the square = l = 0.55 m

length of the diagonal = r = (sqrt2)*l = 0.778 m

initial PE of the system of charges

UA = k*q*q/r + k*q*qo/l + k*q*qo/l

UA = k*q*q/r + 2*k*q*qo/l

UA = (9*10^9*9*10^-6*9*10^-6)/0.778 - (2*9*10^9*9*10^-6*5.4*10^-8)/0.55

UA = 0.937 - 0.016 = 0.921 J

at point C

distance of qo from q = r1 = r/2 = 0.389 m

Uc = k*q*q/r + k*q*qo/r1 + k*q*qo/r1

UA = k*q*q/r + 2*k*q*qo/r1

UA = (9*10^9*9*10^-6*9*10^-6)/0.778 - (2*9*10^9*9*10^-6*5.4*10^-8)/0.389

UA = 0.937 - 0.022 = 0.915 J

change in energy = dU = 0.006 J

from work energy theorem

dU = dKE = 0.5*m*vo^2

vo = sqrt(2*dU/m)

vo = sqrt((2*0.006)/(1.1*10^-8))

vo = 1.0445*10^3 m/s   <---------answer

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