We move an electron from an infinite distance to a point at distance R = 8.00 cm from a tiny charged ball. The move requires work W = 2.16 × 10–13 J by us. (a) What is the charge Q on the ball? In Fig. 24-61b, the ball has been sliced up and the slices spread out so that an equal amount of charge is at the hour positions on a circular clock face of radius R = 8.00 cm. Now the electron is brought from an infinite distance to the center of the circle. (b) With that addition of the electron to the system of 12 charged particles, what is the change in the electric potential energy of the system?

Explanation:

(a) the work done results in a potential energy gain

       From Workdone, W = qΔV

            = (-e)( Q/4πε0R)

Given:

distance, R = 8.00cm= 0.08m

workdone, W = 2.16 * 10⁻¹³ J

we are to find the charge, Q on the ball    

W= (-e) 9 * 10⁹ Q/R

making Q subject of the formula,

 Q = WR/[(-e) * 9* 10⁹]

     = - (2.16 * 10⁻¹³ * 0.08)/[9* 10⁹  * 1.6 * 10⁻¹⁹] = 1.2*10⁻⁵ C

(b)

      the quantity of work is the same so increase in potential energy is ΔU = 2.16 * 10⁻¹³ J

a) The charge Q is -1.2x10⁻⁵C

b) The change in the electric potential energy is 2.16x10⁻¹³J

Explanation:

Please, the solution is in the Word file attached

acceleration= v2-v1/t so 8m/s-3m/s / 3s =to 8/3 or

2.67 m/s^2

hope i

-lavira: )