Question

A charge of 2.50 μC is held fixed at the origin. A second charge of 2.50 μC is released from rest at the position (1.25 m, 0.570 m).

* If the mass of the second charge is 2.48 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

potential due to a point charge q at distance r,

V = k q / r

at the given location,

Vi = (9 x 10^9)(2.50 x 10^-6) / sqrt(1.25^2 + 0.570^2)

= 16377.6 Volt

when it is at infinity. Vf = 0

delta(KE) = - q (Vf - Vi)

(2.48 x 10^-3) (v^2 - 0)/2 = (2.50 x 10^-6) (16377.6)

v = 5.75 m/s .........Ans

for that applying energy conservation,

PEi + KEi = PEf + KEf

{ C = k (2.50 uC) (2.50 uC) }

C / sqrt(1.25^2 + 0.570^2) + 0 = C / r + (C / 4 x sqrt(1.25^2 + 0.570^2))

C / 1.37 = C / r + C / (5.5)

1/1.37 = 1/r + 1/5.5

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