Four identical capacitors are connected in series with a 10 V battery as shown in the figure. Potentials at A and B are

Let q = charge on each capacitorqC+qC+qC+qC=10 ⇒     qC=104 ⇒      qC=2.5 ∴     VA−VN=3qC                         =3×2.5=7.5 V          VA−0=7.5V ⇒     VA=7.5V again    VN−VB=qC ⇒     0−VB=2.5V ⇒      VB=−2.5V