Question

Two charges 6 micro coulomb and 15 micro coulomb are placed along the x- axis at x = 0 and x = 2mrespectively. A negative charge is placed between them such that the resultant force on it is zero. The negative charge is placed at

Answer 1

$$\begin{gathered} \mathrm{Electric}\mathrm{field}\mathrm{due}\mathrm{to}\mathrm{point}\mathrm{charge}: \\ \mathrm{E}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2} \\ \mathrm{Condition}\mathrm{for}\mathrm{zero}\mathrm{force}: \\ \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1}{{\mathrm{x}}^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_2}{{\left(2-\mathrm{x}\right)}^2} \\ \frac{{\mathrm{q}}_1}{{\mathrm{x}}^2}=\frac{{\mathrm{q}}_2}{{\left(2-\mathrm{x}\right)}^2} \\ {\left(\frac{2-\mathrm{x}}{\mathrm{x}}\right)}^2=\frac{{\mathrm{q}}_2}{{\mathrm{q}}_1} \\ \frac{2-\mathrm{x}}{\mathrm{x}}=\sqrt{\frac{15}{6}} \\ \frac{2-\mathrm{x}}{\mathrm{x}}=1.58 \\ 2-\mathrm{x}=1.58\mathrm{x} \\ \mathrm{x}=\frac{2}{2.58} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{78}\mathbf{}\mathbf{m}\mathbf{}\mathbf{(}\mathbf{from}\mathbf{}\mathbf{6}\mathbf{}\mathbf{micro}\mathbf{}\mathbf{coulomb}\mathbf{)} \end{gathered}$$