Please see and answer BOTH questions showing work!! Thanks! Map Sapling Learning

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Please see and answer BOTH questions showing work!! Thanks!

Map Sapling Learning macmillan learning A 18.41-mC charge is placed 31.65 cm to the left of a 89.73-mC charge, as shown in the figure, and both charges are held stationary. A particle with a charge of -7.051 LC and a mass of 28.81 g (depicted as a blue sphere) is placed at rest at a distance 28.49 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 VA at that point? -7.051 Number m/ s 28.49 cm +18.41 mC IK- 9.495 cm +89.73 mC 31.65 cm Previous Give Up & View SolutionCheck AnswerNextExit Hint

Explanation / Answer

given
q1 = 18.41 mC
d = 31.65 cm
q2 = 89.73 mC
q3 = -7.051 uC
m = 28.81 g
r = 28.49 cm

let speed of particel at point A be va

then
from conservation of energy

kq1q3/sqrt(d^2 + r^2) + kq2q3/r = kq1q3/9.495cm + kq2q3/(d - 9.495cm) + 0.5mva^2

8.98*10^11*(-7.051*10^-9)(18.41/sqrt(31.65^2 + 28.49^2) + 89.73/28.49) = 8.98*10^11*(-7.051*10^-9)(18.41/9.495 + 89.73/22.155) + 0.5*0.02881*va^2
va = 1028.6324 m/s

given, m = 9.05*10^-14 kg
q1 = -2.7 pC
q2 = -2.9 pC
r = 3.75*10^-6 m

hence initial speed = v
so from conservation of energy
mv^2 = kq1q2/2r
hence
v = 321.85786 m/s

maximum acceleration = kq1q2/m*(2r)^2 = 1.3812331*10^10 m/s/s

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