Question

Consider the system of capacitors shown in the figure below (C₁ = 7.00

Answer 1

a. Here C1 & 6uF are in series and so are C2 & 2uF. And then equaivalent of those two are in parallel.

Ceq = (C1\6uF) || (C2\2uF) ... \ means series

{Series... 1/Ceq = 1/Ca + 1/Cb or Ceq = (Ca*Cb) / (Ca+Cb)

Parallel ... Ceq = Ca + Cb}

Ceq = (7\6) || (5\2) = (7*6)/(7+6) + (5*2)/(5+2) = 42/13 + 10/7 = 424/91 = 4.65uF

Qtotal = CV = 4.65 * 90 uC = 419.34 uC

b. V across C1 & 6uF = +90V

Q1 = C1eq * V = [(6*7)/(6+7)] * 90 = 290.77uC

Q2 = Qtotal - Q1 = C2eq * V = 128.57uC

Nor charge on C1 & 6uF = Q1 = 290.77uC

{How? Say charge on right plate of C1 = +q, then on right plate of C1 it will be -q. Consider the two inner plates i.e. right plate of C1 & left plate of 6uF. They combinedly must be electrically neutral. Hence charge on left plate of 6uF = +q, which is same as that of C1) }

Similarly charge on C2 and 2 uF is same and = 128.57uC

c. V across C1 = Q₁ / C1 = 41.53 V

V across C2 = = Q₂ / C2 = 25.71V

C across 6uF =  Q₂ / 6 = 90 - 41.53 =

C across 2uF =  Q₂ / 2 = 90 - 25.71 = 64.29V

d. Total Energy = 1/2 * Ceq * V² = 1/2 Q_total²/ V = 18.833 mJ

Cheers

PS - Ask if you have any doubt.