A 9 mu F capacitor and a 25 mu F capacitor are connected in

A 9- mu F capacitor and a 25- mu F capacitor are connected in series, and the combination is charged to a potential difference of 22 V. How much energy is then stored in this capacitor combination? Two charges q_1 = + 5 Times 10^9 C and q_2 = + 8 Times 10^-9 C are placed 40.0 cm apart. A third particle of mass 30 mu g, and charge q_3 = - 5 Times 10^-9 C is placed between the two particles such that it is 25 cm from q_1 as shown in the diagram below. Determine the acceleration of of the chage q_3 as a result of the net force applied by the two other charges

Solution

5)

equivalent capacitor , Ceqv = C1*C2/C1+C2 = 9*25/9+25 = 6.617 micro f

Now energy stored in capacitor = 1/2 CV^2 = 1/2 * 6.617 *22^2 = 1601 micro joule = 1.6 mJoule

6) Force due to q1 = Kq1q3/(25*10^-2)^2 towards q1

force due to q2 = kq2q3/(15*10^-2)^2 toward q2

Fo net force = kq3(q1/.25^2 0 q2/.15^2) = 9*10^9 *5*10^-9 (5/.25^2-8/.15^2) * 10^-9 = 12400 * 10^-9 towards q2

so acceleration = 12400 *10*-9 /30*10^-6 = .413 m/s^2