Physics - Electrostatics Question with Solution | TestHub

Potential at point$P$due to charges$- q, 3q,$and$+ q$is given by

Potential at$P$due to$(+ q)$charge + potential at$P$due to$(- q)$charge + potential at$P$due to$(+3q)$charge

i.e.$V_{Net}=V_A+V_B+V_C$

$\left[\mathrm{A}\mathrm{s}\mathrm{ }\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{ }\mathrm{f}\mathrm{i}\mathrm{g}\mathrm{u}\mathrm{r}\mathrm{e} V_C+V_B=0, \because V_C=\frac{Kq}{\sqrt{2}\frac{a}{2}} \because V_B=\frac{{-K}_q}{\sqrt{2}\frac{a}{2}}\right]$

So,$V_{Net}=V_A=\frac{3Kq}{AP}=\frac{3KQ}{\frac{\sqrt{2a}}{2}}$

$\left[\because AP=\sqrt{a^2-{\left(\frac{\sqrt{2a}}{2}\right)}^2}=\sqrt{a^2-\frac{2a^2}{4}}=\frac{\sqrt{2a^2}}{4}=\frac{\sqrt{2}a}{2}\right]$

$V_{Net}= \frac{3q\times 2}{4\pi {\epsilon }_0\sqrt{2}a}$

$\Rightarrow \frac{3q}{2\sqrt{2}\pi {\epsilon }_{0}a}$