Physics - Electrostatics Question with Solution | TestHub

Let the point $\mathrm{P}$ where resultant field is zero be $x\mathrm{m}$ from $4\mu \mathrm{C}$ charge and $(1-\mathrm{x})\mathrm{m}$ distance apart from $-2\mu \mathrm{C}$ charge. Since field is zero at this point then, $\begin{array}{l}\overrightarrow{\mathrm{E}}={\overrightarrow{\mathrm{E}}}_1+{\overrightarrow{\mathrm{E}}}_2=0\\ \Rightarrow |\overrightarrow{\mathrm{E}}|=\frac{1}{4\pi {ϵ}_0}\frac{{\mathrm{q}}_1}{{\mathrm{r}}_1^2}+\frac{1}{4\pi {ϵ}_0}\frac{{\mathrm{q}}_2}{{\mathrm{r}}_2^2}\\ \Rightarrow 0=\frac{1}{4\pi {ϵ}_0}\left[\frac{4\mu \mathrm{C}}{{\mathrm{x}}^2}+\frac{(-2\mu \mathrm{C})}{(1-\mathrm{x}{)}^2}\right]\\ \Rightarrow \frac{4\mu \mathrm{C}}{{\mathrm{x}}^2}=\frac{2\mu \mathrm{C}}{(1-\mathrm{x}{)}^2}\Rightarrow \frac{2}{{\mathrm{x}}^2}=\frac{1}{(1-\mathrm{x}{)}^2}\\ \Rightarrow 2(1-\mathrm{x}{)}^2={\mathrm{x}}^2\end{array}$ Taking root $\sqrt{2}(1-x)=x$ $\begin{array}{l}\Rightarrow 1.414(1-x)=x\Rightarrow 1.414-1.414x=x\\ \Rightarrow 1.414=(1+1.414)x\Rightarrow x=\frac{1.414}{2.414}\\ \Rightarrow x=0.58\mathrm{m}\end{array}$