Physics - Electrostatics Question with Solution | TestHub

At$\mathrm{P}$there are two fields

(1) Due to$\mathrm{A}={\mathrm{E}}_{\mathrm{A}}=\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}$

(2) Due to$\mathrm{B}={\mathrm{E}}_{\mathrm{B}}=\frac{\mathrm{k}\mathrm{q}}{{\mathrm{r}}^2}$

${\text{r =(a}}^{\text{2}}+{\text{ d}}^{\text{2}}{\text{)}}^{1/2}$at angle$\text{180}-\text{2\hspace{0.17em}θ}$

${\text{E}}_{\text{net}}={\text{ 2E}}_{\text{A}}\text{ cos\hspace{0.17em}}\frac{\text{1}}{\text{2}}\text{ (180}-\text{2θ)\hspace{0.17em}}={\text{ 2E}}_{\text{A}}\text{cos(90 }-\text{\hspace{0.17em}θ)}$

${\text{=2E}}_{\text{A}}\text{\hspace{0.17em}sin\hspace{0.17em}θ =}\frac{\text{2kq}}{{\text{r}}^{\text{2}}}\frac{\text{d}}{\left(\text{r}\right)}$

$\text{=}\frac{\text{2kqd}}{{{\text{(a}}^{\text{2}}{\text{+d}}^{\text{2}}\text{)}}^{3/2}}$in$-\mathrm{z}$direction.