Question

A charge of 3.06 is held fixed at the origin. A second charge of 3.06 is released from rest at the position (1.11 m, 0.524 m). If the mass of the second charge is 2.73 g, what is its speed when it moves infinitely far from the origin? At what distance from the origin does the 3.06 charge attain half the speed it will have at infinity?

Answer 1

Using energy conservation
dPE = dKE
PEf - PEi = 0.5*m*v^2 - 0

here, m = 2.73 g = 2.73*10^-3

v = velocity of charge = ??

PEf = 0

PEi = k*q1*q2/r

where, k = 9*10^9

q1 = q2 = 3.06

r = sqrt((1.11-0)^2 + (0.524 - 0)^2)

r = 1.23 m

So, PEi = (9*10^9)*3.06*3.06/1.23

PEi = 6.85*10^10

now,

0.5*m*v^2 = 6.85*10^10

v = sqrt[(2*6.85*10^10)/(2.73*10^-3)]

v = 7.084*10^6 m/sec

Now,  Let at 'R' distance velocity of charge is(v') = v/2 = (7.084*10^6)/2 = 3.542*10^6

by energy conservation,

KEi + PEi = KEf + PEf

here, KEi = 0

KEf = 0.5*m*(v')^2

PEi = 6.85*10^10

PEf = k*q1*q2/R

So,

0 + 6.85*10^10 = k*q1*q2/R + 0.5*m*v'^2

k*q1*q2/R = 6.85*10^10 - -0.5*(2.73*10^-3)*(3.542*10^6)^2

k*q1*q2/R = 5.137*10^10

R = (9*10^9)*3.06*3.06/(5.137*10^10)

R = 1.64 m

I think velocity part has very high value, So check your question, if it have charge in micro-coulomb, if you have then use this solution

Using q1 = q2 = 3.06*10^-6 C
Using energy conservation
dPE = dKE
PEf - PEi = 0 - 0.5*m*v^2

here, m = 2.73 g = 2.73*10^-3

v = velocity of charge = ??

PEf = 0

PEi = k*q1*q2/r

where, k = 9*10^9

q1 = q2 = 3.06*10^-6 C

r = sqrt((1.11-0)^2 + (0.524 - 0)^2)

r = 1.23 m

So, PEi = (9*10^9)(3.06*10^-6)(3.06*10^-6)/1.23

PEi = 0.0685

now,

0.5*m*v^2 = 0.0685

v = sqrt[(2*0.0685)/(2.73*10^-3)]

v = 7.084 m/sec

Now,  Let at 'R' distance velocity of charge is(v') = v/2 = 7.084/2 = 3.542

by energy conservation,

KEi + PEi = KEf + PEf

here, KEi = 0

KEf = 0.5*m*(v')^2

PEi = 0.0685

PEf = k*q1*q2/R

So, k*q1*q2/R - 0.0685 = 0 - 0.5*m*v'^2

k*q1*q2/R = 0.0685 - 0.5*(2.73*10^-3)*(3.542)^2

k*q1*q2/R = 0.0514

R = (9*10^9)(3.06*10^-6)(3.06*10^-6)/(0.0514)

R = 1.64 m

Please upvote.

So answer of part B will be same in both case, but check which velocity is required. And let me know which one works.