Let the charge on C = q
So, net force on c is equal to zero
So, $\text{F}_{\overline{\text{AC}}}+\text{F}_{\overline{\text{BA}}}=0,$
But $\text{F}_{\overline{\text{AC}}}=\text{F}_{\overline{\text{BC}}}$
$\Rightarrow\frac{\text{kqQ}}{\text{x}^2}=\frac{\text{k}2\text{qQ}}{(\text{d}-\text{x})^2}$
$\Rightarrow2\text{x}^2=(\text{d}-\text{x})^2$
$\Rightarrow\sqrt{2}\text{x}=\text{d}-\text{x}$
$\Rightarrow\text{x}=\frac{\text{d}}{\sqrt{2}+1}$
$=\frac{\text{d}}{\big(\sqrt{2}+1\big)}\times\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}-1\big)}$
$=\text{d}\big(\sqrt{2}-1\big)$
For the charge on rest, $\text{F}_\text{AC}+\text{F}_\text{AB}=0$
$(2.414)^2\frac{\text{kqQ}}{\text{d}^2}+\frac{\text{kq}(2\text{q})}{\text{d}^2}=0$
$\Rightarrow\frac{\text{kq}}{\text{d}^2}\big[(2.414)^2\text{Q}+2\text{q}\big]=0$
$\Rightarrow2\text{q}=-(2.414)^2\text{Q}$
$\Rightarrow\text{Q}=\frac{2}{-\big(\sqrt{2}+1\big)^2}\text{q}$
$=-\Big(\frac{2}{3+2\sqrt{2}}\Big)\text{q}$
$=-(0.343)\text{q}$
$=-\big(6-4\sqrt{2}\big)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
