Question

What is the electric potential at the center of the triangle in figure.

Answer 1

$\because$ Electric potential is a sealer quantity

$\therefore$ It can be added algebraically.

$\Rightarrow V_{net}=V_1+V_2+V_3$

$\Rightarrow V_{net}=\frac{k{q}_1}{r}+\frac{k{q}_2}{r}+\frac{k{q}_3}{r}$

$\Rightarrow V_{net}=\frac{k}{r}(q_1+q_2+q_3)$

$\Rightarrow V_{net}=\frac{9\times {10}^9}{(\frac{20}{\sqrt{3}}\times 100)}(+4+4-4)\times {10}^{-6}$

$=\frac{9\times {10}^9\times 4\times \sqrt{3}\times 100\times {10}^{-6}}{20}$

$V_{net}=3.078\times {10}^{5}V$