Two identical balls having like charges and p | vedclass

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Question

$\mathrm{F}_{1}=\frac{\mathrm{k} \mathrm{Q}_{1} \mathrm{Q}_{2}}{\mathrm{r}^{2}} \quad \mathrm{F}_{2}=\frac{\mathrm{k}\left(\mathrm{Q}_{1}+\mathrm{Q}_{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\mathrm{F}_{1}=\frac{\mathrm{k} \mathrm{Q}_{1} \mathrm{Q}_{2}}{\mathrm{r}^{2}} \quad \mathrm{F}_{2}=\frac{\mathrm{k}\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right)^{2}}{\mathrm{r}^{2}}$

Given $\quad \mathrm{F}_{2}=4.5 \mathrm{F}_{1}$

${\frac{\mathrm{k}\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right)^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{k} \mathrm{Q}_{1} \mathrm{Q}_{2}}{\mathrm{r}^{2}} 4.5} $

${\mathrm{Q}_{1}^{2}+\mathrm{Q}_{2}^{2}=2.5 \mathrm{Q}_{1} \mathrm{Q}_{2}}$

${\left(\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}}\right)^{2}+1=\frac{2.5 \mathrm{Q}_{1}}{\mathrm{Q}_{2}}}$

Let $\frac{Q_{1}}{Q_{2}}=x$

$\Rightarrow x^{2}+1-2.5 x=0$

$\Rightarrow x^{2}+1-2.5 x=0$

$\quad x^{2}-2 x-0.5 x+1=0$

$(X-2)(x-0.5)=0$

$x=2$ or $\frac{1}{2}$

$\Rightarrow \frac{Q_{1}}{Q_{2}}=2$

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