Capacitance

Q15 ans: (a) 50/3 uV at each point
(b) zero

(a) Capacitor = (4 x 8)/(4 + 8) = 8/3μ and 6 x 3/6 x 3 = 2μF (i) The charge on the capacitance 8/3μF
∴ Q = 8/3 x 50 = 400/3
∴ The potential at 4μF = 400/3 x 4 = 100/3 at 8μF = 400/3 x 8 = 100/6 The Potential difference = 100/3  100/6 = 50μV
(ii) Hence the effective charge at 2μF = 50 × 2 = 100μF ∴ Potential at 3μF =100/3;
Potential at 6μF = 100/6
∴ Difference = 100/3  100/6 = 50/3μV
∴ The potential at C & D is 50/3μV
(b) ∴ p/q = R/S = 1/2 = 1/2 = it It is balanced. So from it is cleared that the wheat star bridge balanced. So the potential at the point C & D are same.
So no current flow through the point C & D. So if we connect another capacitor at the point C & D the charge on the capacitor is zero

@yashkapoor the potential difference of upper two capacitors us 50uV so how is the potential at point C 50/3 uV