Trajectory in an arbitrary electrostatic field

Consider every infinitesimal arc that makes up the trajectory. If the heavier mass can follow the same path for each and every infinitesimal arc, then the heavier mass can then follow the exact same trajectory all the way. As the arcs are infinitesimal, we can approximate the velocity of the masses in each arc to be constant and to change only between arcs. For an arc of length dl and radius of curvature r, the lighter mass enters at speed v, meaning the radial force it experience is $mv^2/r$ . For the heavier mass with 7 times the charge, the radial force it experience is $F=7mv^2/r$ . Since the heavier mass follows the same arc of length dl and radius of curvature, we have $F = 10mV^2/r$ . Hence V of the heavier mass is $\sqrt{7/10}v$ . To check for consistency, we need to also inspect the tangential forces, which can also be assumed to be constant over the infinitesimal arc. A tangential force of f over dl will do work of f~dl. The change of velocity for the lighter mass, dv, can be found using the following equation $f~dl = (1/2)m(v^2 -(v-dv)^2)$ . Solving, $dv=f*dl/(mv)$ . A tangential force of 7f applies to the heavier mass. Hence the change of velocity for the heavier mass, dV, can be found using the following equation $7f~dl = (1/2)10m(V^2 -(V-dV)^2)$ . Solving, $dV=7f~dl/(10mV)$ . Hence, $V-dV = \sqrt{7/10}(v-dv)$ . Thus $V = \sqrt{7/10}v$ at all times along the trajectory and hence the heavier mass will follow the same path at all times, given that the initial velocity of the heavier mass is $\sqrt{7/10} = 0.8367$ times of the lighter one.

The second particle is 10 times heavier with 7 times more charge, and therefore experiences 0.7 times the acceleration of the first particle at any given location. We do not wish the path to change so distance must be invariant, but we can rescale time to account for the difference in acceleration. Since this acceleration is measured in $m/s^2$ and velocity is measured in $m/s$ , a reduction in velocity of $\sqrt{0.7} \approx 0.8367$ produces the necessary adjustment of 0.7 in acceleration.

Note that $\vec{\tau}=\frac{v}{v}$ is the unit tangent vector to the particle's path at any point. The particle's trajectory can be parametrized by means of the arc-length s (the distance between any point on the curve an the initial point A). We can write $\frac{d}{ds} \frac{\vec{v}}{v}= \frac{1}{v}\frac{d}{dt} \frac{\vec{v}}{v}=\frac{ v \frac{d\vec{v}}{dt}- \vec{v}\frac{dv}{dt}}{v^{3}}.$ From Newton's second law we have that: $\frac{d\vec{v}}{dt}= \frac{q}{m} \vec{E}.$ Multiplying both sides by $\vec{v}$ we obtain: $\vec{v}\cdot \frac{d\vec{v}}{dt}=\frac{q}{m} \vec{E}\cdot \vec{v} \rightarrow \frac{dv}{dt}= \frac{q}{m} \vec{E}\cdot \vec{ \tau}.$ Here we have used the fact that $\vec{v}\cdot \frac{d \vec{v}}{dt}= v \frac{dv}{dt}$ . Moreover, from conservation of energy we have that $\mathcal{E}=\frac{mv^{2}}{2}+ q V=\textrm{constant.}$

Using the above relations we arrive at $\frac{d}{ds} \frac{\vec{v}}{v}= \frac{\vec{E}(s)- (\vec{E}(s)\cdot \vec{\tau}) \vec{\tau}}{2 (\frac{\mathcal{E}}{q}- V(s))}.$ By knowing the electric field distribution $\vec{E}(s)$ and the potential $V(s)$ one could (in principle) find the shape of the particle's path. Note, however, that the two particles will follow the same path if they have the same ratio $\frac{\mathcal{E}}{q}$ . Therefore, we following relation must hold: $\frac{m v_{1}^{2}}{2 q}+ V_{A}= \frac{M v_{2}^{2}}{2 Q}+ V_{A}$ where $V_{A}$ is the electric potential at point A. Thus $v_{2} = \sqrt{\frac{m Q}{M q}} v_{1} =0.84 m/s.$