Question 4.3: A practical application of electrostatics is in electrostati...

Ignoring the coulombic force between particles, the electrostatic force is acting horizontally while the gravitational force (weight) is acting vertically on the particles. Thus

QE=m\frac{d^{2}x}{dt^{2}}a_{x}

or

\frac{d^{2}x}{dt^{2}}=\frac{Q}{m}E

Integrating twice gives

x=\frac{Q}{2m}Et^{2}+c_{1}t +c_{2}

where c_{1} and c_{2} are integration constants. Similarly

-mg=m\frac{d^{2}y}{dt^{2}}

or

\frac{d^{2}y}{dt^{2}}=-g

Integrating twice, we get

y=-1/2gt^{2}+c_{3}t+c_{4}

Since the initial displacement is zero

x\left(t=0\right)=0\rightarrow c_{2}=0

y\left(t=0\right)=0\rightarrow c_{4}=0

Also, because of zero initial velocity

\frac{dx}{dt}|_{t=0}=0\rightarrow c_{1}=0

\frac{dy}{dt}|_{t=0}=0\rightarrow c_{3}=0

Thus

x=\frac{QE}{2m}t^{2},      y=-\frac{1}{2}gt^{2}

When y=-80cm=-0.8m

t^{2}=\frac{0.8\times 2}{9.8}=0.1633

and

x=1/2\times 9\times 10^{-6}\times 5\times 10^{5}\times 0.1633=0.3673m

The separation between the particles is 2x=73.47cm.