Figure shows three arrangements of electric field lines. In each arrangement, a proton is released from rest at point $\mathrm{P}$ and then accelerated through point $\mathrm{Q}$ by the electric field. Points $\mathrm{P}$ and $\mathrm{Q}$ have equal separations in the three arrangements. If $p_1 p_2$ and $p_3$ are linear momentum of the proton at point $\mathrm{Q}$ in the three arrangement respectively, then

The work done by the electric field on a proton moving from point P to point Q is given by $W = \int_{P}^{Q} \vec{F} \cdot d\vec{r} = q \int_{P}^{Q} \vec{E} \cdot d\vec{r}$.

By the work-energy theorem, the change in kinetic energy is equal to the work done, so $\Delta K = K_Q - K_P = W$. Since the proton is released from rest at P, $K_P = 0$, thus $K_Q = W = \dfrac{p^2}{2m}$, which implies $p = \sqrt{2m W}$.

The work done $W$ is equal to $q \Delta V$, where $\Delta V$ is the potential difference between P and Q. The potential difference is $\Delta V = \int_{P}^{Q} \vec{E} \cdot d\vec{r}$.

In arrangement 1, the electric field lines are parallel and uniformly spaced, indicating a uniform electric field. The potential difference is $\Delta V_1 = E \cdot d$.

In arrangements 2 and 3, the electric field lines are non-uniform. The density of field lines represents the magnitude of the electric field. In both arrangements 2 and 3, the field lines are closer together near Q compared to P, meaning the average electric field strength along the path from P to Q is higher than in arrangement 1.

Comparing the configurations, the integral of the electric field along the path from P to Q is greater in arrangements 2 and 3 than in arrangement 1. Since the field lines are identical in density and distribution between P and Q for arrangements 2 and 3, the potential difference $\Delta V_2 = \Delta V_3 > \Delta V_1$.

Therefore, the momentum $p = \sqrt{2mq \Delta V}$ satisfies $p_1 < p_2 = p_3$.

Answer: $p_1&lt; p_2=p_3$