Question

An electric dipole of moment $$\vec{p}=(-\hat{i}-3\hat{j}+2\hat{k})\times 10^{-29}$$ C.m is at the origin $$(0, 0, 0)$$. The electric field due to this dipole at $$\vec{r}=+\hat{i}+3\hat{j}+5{k}$$ (note that $$\vec{r}\cdot \vec{p}=0$$ is parallel to):

Answer 1

Correct option is C. $$(+\hat{i}+3\hat{j}-2\hat{k})$$

Refer above image.

$$\vec{P}=(-\hat {i }-3\hat { j }+2\hat { k})\times 10^{-29}cm$$

position vector $$\vec{r}=\hat { i }+3\hat { j }+5\hat { k }$$

$$\vec{p}.\vec{r}=(-\hat { i}-3\hat { j }+2\hat { k }).(\hat { i }+3\hat { j }+5\hat { k })$$

$$=-1-9+10=0$$

$$\therefore$$ Vectors $$\vec{P}$$ and $$\vec{r}$$ are perpendicular to each other.

From the figure, electric field at $$\vec{r}$$ will be anti-parallel to the direction of $$\vec{p}$$ (The given point is an axial point and electric field at an axial point is anti-parallel to the direction of $$\vec{p}$$)

$$\therefore \vec{E}=-\lambda\vec{p}$$ where $$\lambda > o$$

$$=-\lambda(- \hat { i }-3 \hat { j }+2 \hat { k })\times 10^{-29}$$

$$=\lambda( \hat { i }+3 \hat { j }-2 \hat { k })\times 10^{-29}$$

$$\therefore \vec{E}$$ is parallel to $$ \hat { i }+3 \hat { j }-2 \hat {k }$$

Option (C) is correct.