Physics - Electrostatics Question with Solution | TestHub

$E=\frac{Kq}{r^2}\cdot \hat{r}=\frac{Kq}{r^3}\cdot r$ Here, $\begin{array}{rl}r& =\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}\\ & =\sqrt{(30{)}^2+(30{)}^2+0^2}\\ & =30\sqrt{2}\mathrm{cm}\\ & =30\sqrt{2}\times 10^{-2}\mathrm{m}\end{array}$ $\text{ and }q=8\times 10^{-3}\times 10^{-6}\mathrm{C}$ Also, $\mathbf{r}=(30\hat{\mathbf{i}}+30\hat{\mathbf{j}})\times 10^{-2}\mathrm{m}$ $\begin{array}{rl}\text{ So, }\mathrm{E}& =\frac{9\times 10^9\times 8\times 10^{-3}\times 10^{-6}}{{\left(30\sqrt{2}\times 10^{-2}\right)}^3}\times (30\hat{\mathbf{i}}+30\hat{\mathbf{j}})\times 10^{-2}\\ & =\frac{9\times 8\times 10^9\times 10^{-11}}{27\times 2\sqrt{2}\times 10^{-6}\times 10^3}\times 30(\hat{\mathbf{i}}+\hat{\mathbf{j}})\\ & =\frac{9\times 8\times 10^2\times 3}{27\times 2\sqrt{2}}(\hat{\mathbf{i}}+\hat{\mathbf{j}})\\ & =200\sqrt{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}}){\mathrm{NC}}^{-1}\end{array}$