Physics - Electrostatics Question with Solution | TestHub

P is origin. We can see from diagram electric field at P is zero but we have to check if it is zero at any other point x -axis we have to give particle just as much velocity to reach to point where electric field is zero. (Called neutral point) Let field is zero at $\theta$. ${\left({\mathrm{E}}_{\theta }\right)}_{\text{Net }}=0$ $2\mathrm{E}\cos \theta =2{\mathrm{E}}_0\cos {\theta }_1$ $2\times \frac{\mathrm{kq}}{{\mathrm{r}}^2}\times \frac{\mathrm{x}}{\mathrm{r}}=2\times \frac{\mathrm{k}8{\mathrm{q}}_0}{{\mathrm{r}}_1^2}\times \frac{\mathrm{x}}{{\mathrm{r}}_1}$ $\frac{1}{r^3}=\frac{8}{r_1^3}\Rightarrow \frac{r_1}{r}=2$ For $\mathrm{Q}\frac{{\mathrm{r}}_1}{\mathrm{r}}=+2$ ${\mathrm{r}}_1=2\mathrm{r}$, again $\mathrm{rl}=4$ $\sqrt{x^2+\frac{27}{2}}=2\sqrt{x^2+\frac{3}{2}},\mathrm{r}=2$ ${\mathrm{x}}^2+\frac{27}{2}=4{\mathrm{x}}^2+\frac{12}{2};3{\mathrm{x}}^2=\frac{15}{2}\Rightarrow \mathrm{x}=\sqrt{\frac{5}{2}}$

To reach at origin particle has to reach at Q the it will be attracted towards origin by electric fields of $-1\mu \mathrm{C}$ Apply work energy theorem from $\infty$ to Q . $-\mathrm{q}\Delta \mathrm{V}={\mathrm{k}}_{\mathrm{f}}-{\mathrm{k}}_{\mathrm{i}}$ $-q\left(v_Q-v_{\infty }\right)=0-\frac{1}{2}m{v}_0^2$ $-1\times 10^{-6}\left[\left(\frac{2\times 8\times 10^{-6}\times 9\times 10^9}{r_1}-\frac{2\times 10^{-6}\times 9\times 10^9}{r}\right)-0\right]=-\frac{1}{2}m{v}_0^2$ $-10^6\times 2\times 10^{-6}\times 9\times 109\left(\frac{8}{4}-\frac{1}{2}\right)=-\frac{1}{2}{\mathrm{mv}}_0^2$ Solving we get ${\mathrm{v}}_0=3\mathrm{m}/\mathrm{s}$