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

Sometimes, the most intimidating physics problems are just testing your ability to read between the lines. When you first look at this question, you see a messy dipole moment vector $\mathbf{p}=(-\hat{\mathbf{i}}-3\hat{\mathbf{j}}+2\hat{\mathbf{k}})\times 10^{-29}$ C-m and an equally messy position vector $\mathbf{r}=+\hat{\mathbf{i}}+3\hat{\mathbf{j}}+5\hat{\mathbf{k}}$. Your first instinct might be to panic and start plugging these into the general 3D formula for the electric field of a dipole.

But wait! The question gives you a beautiful, elegant gift in parentheses: $\mathbf{r}\cdot \mathbf{p}=0$.

What does a dot product of zero actually mean in the physical world? It is the mathematical universe's way of telling you that two vectors are perfectly perpendicular to each other.

If the position vector $\mathbf{r}$ is perpendicular to the dipole moment $\mathbf{p}$, it means the point we are investigating lies exactly on the equatorial plane of the dipole.

Now, let's recall the behavior of electric fields around a dipole. On the axial line, the electric field points in the same direction as the dipole moment. But on the equatorial plane, the story is different.

If you place a positive test charge anywhere on the equatorial plane, it gets pushed away by the positive charge of the dipole and pulled in by the negative charge. Because the point is equidistant from both charges, the vertical components of these forces cancel out perfectly. The horizontal components add up, pointing exactly from the positive charge towards the negative charge.

Since the dipole moment vector $\mathbf{p}$ points from negative to positive, the resulting electric field $\mathbf{E}$ must point in the exact opposite direction. Mathematically, we say $\mathbf{E}$ is anti-parallel to $\mathbf{p}$.

Because we only care about the direction of the electric field, we don't need to calculate the messy magnitudes involving $1/r^3$. We simply know that the direction of $\mathbf{E}$ is proportional to $-\mathbf{p}$.

Let's execute the final step:

Distributing the negative sign, we get our final answer:

And just like that, a seemingly complex 3D vector problem collapses into a single, elegant conceptual realization.