Question

A charge of 2.87 μC is held fixed at the origin. A second charge of 3.60 μC is released from rest at the position (1.25 m, 0.570 m).

If the mass of the second charge is 2.29 g , what is its speed when it moves infinitely far from the origin?

At what distance from the origin does the second charge attain half the speed it will have at infinity?

Answer 1

PE = k q1 q2 / r

Applying energy conservation,

PEi + KEi = PEf + KEf

(9 x 10^9 x 2.87 x 10^-6 x 3.60 x 10^-6)/sqrt(1.25^2 + 0.570^2) + 0 = 0 + (2.29 x 10^-3) v^2 / 2 + 0

( when charge is at very far away then PE = 0 )

v = 7.7 m/s ....Ans

for v = 7.7/2 = 3.8 m/s

putting in ,

(9 x 10^9 x 2.87 x 10^-6 x 3.60 x 10^-6)/sqrt(1.25^2 + 0.570^2) + 0 = (9 x 10^9 x 2.87 x 10^-6 x 3.60 x 10^-6 / r ) + (2.29x 10^-3 x 3.8^2 / 2)

0.0677 + 0 = 0.092988 / r + 0.0165338

r = 1.82 m ...........Ans