$V_{A} - V_{B}$ of given circuit is [in volts]

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Solution

Let potential of B is 0 and charge on 4th capacitor be Q then potential of l is $\frac{- Q}{2}$ then potential of P is 1.5 and that of m is $1.5 \frac{- Q}{2}$
So charge on 3rd capacitor is $Q_{3} = 2 [v_{p} - v_{m}] Q_{3} = 2 [1.5 - 1.5 + \frac{Q}{2}] = Q$
In same way $Q_{2} = Q_{1} = Q$
Only possible value of Q is zero.
So $V_{A B}$ = 1.5 + 1.5 + 1.5 = 4.5

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