# Direction of Force on a Dipole in a given Electric field

How should I determine the direction of the force in the given case: (p is the electric dipole moment)

Using the following formula

My book has the following :

I am unable to understand the given correlation between the direction of Force and the formula. I am confused with the gradient of E and it's dot product (correct me if I am wrong) with p .

• Please don't close the question as it is not any homework problem. I just need some help with the formula. Commented May 15, 2020 at 21:20
• Is this a high-school book? $p$ is supposed to be a vector. It’s bizarre that $\mathbf F$ and $\mathbf E$ are but $p$ isn’t. What is $l$ supposed to be? Commented May 15, 2020 at 21:45
• Yes it's a high school book. I think p instead of P is a printing error. Idk about l but does it matter? We have the p vector and I am just asking for direction of Force. Commented May 15, 2020 at 22:19
• $l$ has to be the distance along the direction of $\mathbf p$. Commented May 15, 2020 at 22:19
• I’m used to seeing $\mathbf F=(\mathbf p\cdot\nabla)\mathbf E$. Commented May 15, 2020 at 22:21

take advantage of the fact that $$\vec p$$ is a vector.
Resolve $$\vec p$$ into two components - $$\vec p_1$$ and $$\vec p_2$$ one along the line joining charge $$q$$ and the centre of dipole and one perpendicular to it (ref fig 1). the
Electric field at an axial point of dipole : $$\vec E_{axial}= \frac{1}{4\pi\epsilon_0}\frac{2p_2}{r^3}$$
Electric Field along the equitorial plane : $$\vec E_{eq}= \frac{1}{4\pi\epsilon_0}\frac{p_1}{r^3}$$
Now, the net electric field at $$q$$ is $$\vec E=\vec E_{eq}+\vec E_{axial}$$ and force on $$q$$ is just $$q\vec E$$
Think of a “finite-size dipole” as a positive charge at the tip of the $$\mathbf p$$ vector and an equal negative charge at the tail. The forces exerted by $$q$$ on these two charges will not be exactly equal and opposite, because the field of $$q$$ isn’t uniform and the two charges aren’t at the same point. Draw the two force vectors and add them to get the net force on the dipole.