Question

A parallel-plate capacitor has $2.0\mathrm{cm}\times 2.0\mathrm{cm}$electrodes with surface charge densities $\pm 1.0\times {10}^{-6}\mathrm{C}/{\mathrm{m}}^2$. A proton traveling parallel to the electrodes at $1.0\times {10}^6\mathrm{m}/\mathrm{s}$ enters the center of the gap between them. By what distance has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the field is uniform inside the capacitor and zero outside the capacitor.

Short answer

The proton been deflected sideways when it reaches the far edge of the capacitor is

$a=0.108\times {10}^{14}\frac{{\mathrm{m}}^2}{\mathrm{s}}$

$t=2\times {10}^{-8}\sec$

$y=2.2\mathrm{mm}$

Step-by-step solution

Acceleration and Velocity

$\to \eta =1\times {10}^{-6}$( charge densities)

From inside battery, there must be an electric current due to just a paired surface.

$E=\frac{\eta }{{ϵ}_o}=\text{acceleration}$

$a=\frac{qE}{m_p}$

$=\frac{q\eta }{{ϵ}_o{m}_p}$

$a=\frac{1.6\times {10}^{-19}\times {10}^{-6}}{8.85\times {10}^{-12}\times 1.67\times {10}^{-27}}$

$a=0.108\times {10}^{14}\frac{{\mathrm{m}}^2}{\mathrm{s}}$

So there is no force as in downward motion, the pace is same:

$d=vt$

$t=\frac{d}{v}$

$=\frac{2\times {10}^{-2}}{{10}^6}$

$=2\times {10}^{-8}\sec$

Distance

With in vertical motion, ion sweep area

$y=ut+\frac{1}{2}a{t}^2$

$y=\frac{1}{2}\times 0.108\times {10}^{14}\times {\left(2\times {10}^{-8}\right)}^2$

$y=2.2\mathrm{mm}$