As shown in the figure, charges $+\mathrm{q}$ and $-\mathrm{q}$ are placed at the vertices B and C of an isosceles triangle. The potential at the vertex A is

• A

  $\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{2\mathrm{q}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$

• B

  Zero

• C

  $\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{\mathrm{q}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$

• D

  $\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{(-\mathrm{q})}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$