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 capacitor
$$\begin{gathered} \frac{\mathrm{q}}{\mathrm{C}}+\frac{\mathrm{q}}{\mathrm{C}}+\frac{\mathrm{q}}{\mathrm{C}}+\frac{\mathrm{q}}{\mathrm{C}}=10 \\ \Rightarrow \frac{\mathrm{q}}{\mathrm{C}}=\frac{10}{4} \\ \Rightarrow \frac{\mathrm{q}}{\mathrm{C}}=2.5 \\ \therefore {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{N}}=\frac{3\mathrm{q}}{\mathrm{C}} \\ =3\times 2.5=7.5 \mathrm{V} \\ {\mathrm{V}}_{\mathrm{A}}-0=7.5\mathrm{V} \\ \Rightarrow {\mathrm{V}}_{\mathrm{A}}=7.5\mathrm{V} \\ \mathrm{again} {\mathrm{V}}_{\mathrm{N}}-{\mathrm{V}}_{\mathrm{B}}=\frac{\mathrm{q}}{\mathrm{C}} \\ \Rightarrow 0-{\mathrm{V}}_{\mathrm{B}}=2.5\mathrm{V} \\ \Rightarrow {\mathrm{V}}_{\mathrm{B}}=-2.5\mathrm{V} \end{gathered}$$