Yes, it means that the net work done in moving a test charge vertically, along the equitorial line of the dipole is zero.
In fact, take this example -
You have an electric dipole with length $2l$, and let P be any point where a test charge (a unit positive charge) on the equitorial line of the dipole at a distance $r$ from the midpoint $O$ of the dipole is kept. Something like this -
(Pardon me, I found this image on Google; but it does serve the purpose)
As you can see, the net electric field at the point $P$ is only due to the the horizontal components of the electric field due to both the charges, since they add up ($E\cos \theta$), and the vertical components ($E\sin \theta$) due to each charge, cancel out. The net electric field at the equitorial line isn't zero, and so isn't the force. In fact, you can clearly see that the direction of electric field is in the direction where the potential decreases.
Let's find out the potential at a point in the equitorial line and do the math. 531
Since potential is a scalar, it would simply add up.
Hence, potential at the same point P in the equitorial line of the dipole (same scenario as shown in the diagram) is $$ V_{\text{net}} = V_{+q} + V_{-q} $$ $$ V_{\text{net}} = \dfrac{1}{4\pi \epsilon_0} \Bigg(\dfrac{q}{\sqrt{r^2 + l^2}} - \dfrac{q}{\sqrt{r^2 + l^2}}\Bigg) = 0 $$ $$ \implies V_{\text{net}} = 0. $$ What this actually means is that the work done in moving the test charge along the equitorial line, i.e, moving it vertically is zero (Since $W = qV$ and $V = 0 \implies W = 0$). As you can see, $V_{\text{net}} = 0$ does not imply $F_{\text{net}} = 0$. (Since $E_{\text{net}} \neq 0$, as seen from the diagram above) In our case, $r = 0$ doesn't change anything. $V$ still remains $0$.
Work would be done, if you were to move the test charge horizontally, along the axial line of the dipole, but that isn't the case here. 780