Physics - Elecrostatics Question with Solution | TestHub

As $\overrightarrow{r}·\overrightarrow{p}=0$, this means that $\overrightarrow{r}$ is perpendicular to $\overrightarrow{p}$.

For an electric dipole, electric field at equatorial position is antiparallel to $\overrightarrow{p}$.

$\Rightarrow \overrightarrow{E}=-\lambda \overrightarrow{p}$,

where, $\lambda$ is an arbitrary positive constant.

$\Rightarrow \overrightarrow{E}=-\lambda \left(-\hat{\mathrm{i}}-3\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)\times {10}^{-29} \mathrm{C} {\mathrm{m}}^{-2}$

$\Rightarrow \overrightarrow{E}=\lambda \left(\hat{\mathrm{i}}+3\hat{\mathrm{j}}-2\hat{\mathrm{k}}\right)\times {10}^{-29} \mathrm{C} {\mathrm{m}}^{-2}$

For $\lambda =\frac{1}{{10}^{-29}}$ 

$\Rightarrow \overrightarrow{E}=\left(\hat{\mathrm{i}}+3\hat{\mathrm{j}}-2\hat{\mathrm{k}}\right) \mathrm{C} {\mathrm{m}}^{-2}$