As shown in the figure, charges \[+q\] and \[-q\] are placed at the vertices \[B\] and \[C\] of an isosceles triangle. The potential at the vertex A is [MP PET 2000]

\[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]

\[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]

\[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{(-q)}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]

As shown in the figure, charges \[+q\] and \[-q\] are placed at the ve

1.	As shown in the figure, charges \[+q\] and \[-q\] are placed at the vertices \[B\] and \[C\] of an isosceles triangle. The potential at the vertex A is [MP PET 2000]
A.	\[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
B.	Zero
C.	\[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
D.	\[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{(-q)}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]
Answer» C. \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\]