Question 19 of 20 Map sapling learning A 33.91-mC charge is placed 39.15 cm to t

Question

Question 19 of 20 Map sapling learning A 33.91-mC charge is placed 39.15 cm to the left of a 73.23-mC charge, as shown in the figure, and both charges are held s (depicted as a blue sphere) is placed at rest at a distance 35.24 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 A particle with a charge of-3.051 C and a mass of 1931 g 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 vA at that point? Number 3.051 C m/s 35.24 cm +33.91 mC +73.23 mC k-11.75 cm K39.15 cm Previous O Check Answer Next Exit Hint

Explanation / Answer

q1 = 33.91 mc

q2 = 73.23 mc

r12 = 39.15 cm

r23 = 35.24 cm

r13 = sqrt(39.15^2+35.24^2) = 52.67 cm

q3 = -3.051 mc

m3 = 19.31 g

initial potential energy

U1 = k*q1*q2/r12 + k*q2*q3/r23 + k*q1*q3/r13

U1 = ((9*10^9*33.91*10^-3*73.23*10^-3)/0.3915) - ((9*10^9*73.23*10^-3*3.051*10^-6)/0.3524) - ((9*10^9*33.91*10^-3*3.051*10^-6)/0.5267)

U1 = 57078257.08917591 J

K1 = 0

at A

U2 = k*q1*q2/r12 + k*q2*q3/r23 + k*q1*q3/r13

r12 = 39.15 cm

r13 = 11.75 cm

r23 = 39.15-11.75 = 27.4 cm

U2 = ((9*10^9*33.91*10^-3*73.23*10^-3)/0.3915) - ((9*10^9*73.23*10^-3*3.051*10^-6)/0.274) - ((9*10^9*33.91*10^-3*3.051*10^-6)/0.1175)

U2 = 57070467.715425909 J

K2 = 0.5*m*v^2

U1 + K1 = U2 + K2

K2 = U1 - U2

0.5*19.31*10^-3*v^2 = 7789.3737500010002

v = 898.2 m/s <<-answer

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.