A 9- mu F capacitor and a 25- mu F capacitor are connected in series, and the co

Question

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

Explanation / Answer

9 uF and 25 uF are in seies

resultant capacitor C = 9*25/(9+25) = 6.62 uF

voltage V = 22 V

Charge across the combinaion Q = CV= 6.62e-6 *22 = 145.59 uC

energy stored E = CV2/2 = 6.62*222 = 1.6 mJ

B)

Force on Q3(-5 nC) due to Q1(5 nC)

distance = 25 cm

F1 = kQ1Q3/d2 = 8.9e+9*5.0e-9*5.0e-9 /0.252 = 3.56e-6 N

Force on Q3 due to Q2(8.0 nC) distance d = 15 cm

F2 = 8.9*5.0e-9*8.0e-9/0.152 = 15.82 e-6 N

F1 and F2 are in opposite directins as both are +ve charges and atrractve on Q3

net force on Q3 = (15.82-3.56)e-6 =12.26e-6 N

mass of Q3 = 30 ug

acceleration = F/m = 12.26/30 = 0.41 m/s2

                                         =

Energy stored E = CV2/2 = 16.07e-6 *222 /2 = 0.78 mJ

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