I know this is a long question, that\'s why I have offered so many points. Pleas

Question

I know this is a long question, that's why I have offered so many points. Please help, I will rate

Problem 20.35

IP A charge of 19.0?C is held fixed at the origin.

A) If a -7.00?C charge with a mass of 3.70g is released from rest at the position (0.925 m, 1.17 m), what is its speed when it is halfway to the origin? units are in m/s

B) Suppose the -7.00?C charge is released from rest at the point x = 12(0.925m) and y = 12(1.17m). When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part A?

I know it's Greater than the speed found in part A, But why??? please explain

C) Find the speed of the charge for the situation described in part B. units are in m/s

Explanation / Answer

A)

Electric Force between charges , F = Kq1*q2 / r^2

K = 9*10^9 N m^2 / C^2

where r is the distance between the 2 charges

r = sqrt ( x^2 + y^2)

we have q1 = 19 ?C = 19 *10^-6 C

q2 = -7.00?C = -7*10^-6 C

We have acceleration = force / mass = F/m

mass of -7 ?C charge = 3.7 gm = 0.0037 Kg

so, acceleration = 9*10^9 * 19*10^-6 * 7*10^-6 / ( x^2 + y^2) * 0.0037

acceleration , a = 323.5135 / ( x^2 + y^2) = 323.5135 / r^2

a = dV/dt where V is velocity of charge q2

we have velocity , V = dr/dt

so, a = d^2 r / dt^2

so, a = d^2 r / dt^2 = 323.5135 / r^2

d^2 t / dr^2 = r^2 / 323.5135

integrating

dt/dr = r^3 / ( 3* 323.5135) + C = r^3 / 970.5405 + C

at time t = 0 , r = sqrt ( 0.925^2 + 1.17^2) = 1.4914841 m, velocity = dr/dt = 0

so, dr/dt = 1/ ( r^3 / 970.5405 + C )

0 = 1/ (

integrating again

Initially at point ( 0.925 m, 1.17 m) r = sqrt ( x^2 + y^2) = sqrt ( 0.925^2 + 1.17^2) = 1.4914841 m

when it is at halfway r = 1.4914841/2 = 0.745742 m ,

coordinates will be ( 0.925/2 , 1.17/2) = ( 0.4625 , 0.585 )

so

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