QUESTION 4 6 points Save Answer Three identical particles of

QUESTION 4 6 points Save Answer Three identical particles of mass m=5 grams and charge q=3.4 microCoulombs are assembled from infinity to form an equilateral triangle with sides of length d-89 cm as shown in the figure Let W be the work required to assemble the triangle of charges by the following method. Charge A is first nailed down. Charge B is then moved from infinity and nailed down at a distance d from Charge A. Finally Charge C is moved in from infinity and nailed down at the apex of the triangle After the triangle is assembled, the nail holding Charge C is removed and it flies off to infinity - ultimately reaching a speed of vc. Charge A and B are still nailed down. Find vc in meters/sec

Solution

When charge were at very far distance apart then PE of system will be zero.

when come and form traingle then PE will be

PE = 3 [ k (3.4uC) (3.4uC) / (0.089^2) ]

PE = (3 x 9 x 10^9 x 3.4 x 10^-6 x 3.4 x 10^-6 ) / (0.089^2)

PE = 39.40 J

So work done W will be W = 39.40 J

when When C is flown to far apart.

then PE of system will be = [ k (3.4uC) (3.4uC) / (0.089^2) ]

PE = (9 x 10^9 x 3.4 x 10^-6 x 3.4 x 10^-6 ) / (0.089^2) = 13.13 J

Using energy conservation,

initial PE + KE = final PE + KE

39.40 + 0 = 13.13 + ( 0.005 v^2 /2 )

26.27 = 0.005 v^2 / 2

v = 102.51 m/s