Question

Three charges of $+5\mu C$ are located at the comes of an equilateral triangle whose sides are 6 cm long. Find the potential at the midpoint of the base of the of the triangle.

Answer 1

Now for solving total potential at mid point we have to calculate value of r for all charges

We know that for equilateral triangle,

$r_1=\frac{\sqrt{3}a}{2}and{r}_2=r_3=\frac{a}{2}$

here, $a=6cm$

so , Total potential,

$V=\frac{kq}{r_1}+\frac{kq}{r_2}+\frac{kq}{r_3}$

$v=kq(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3})$

putting value of$k,q,r_1,r_2,r_3$ we get,

$V=9\times {10}^9\times 5\times {10}^{-6}\times (\frac{1}{\sqrt{3}\times 0.03}+\frac{1}{0.03}+\frac{1}{0.03})$

$V=15\times {10}^5\times (\frac{1}{\sqrt{3}}+2)$

solving we get

$V=3.866\times {10}^{6}v$