Question

A proton is released from rest at the positive plate of a parallel plate capacitor. The charge per unit area on each plate is σ=1.8e^-7 C/m² , and the plates are separated by a distance of 1.5e^-2 m.

a. What is the speed of the proton when it reaches the negative plate? Solve this problem using conservation of energy.

b.Solve (a) using kinematics equations.

Answer 1

Electric field between the plates of the capacitor $E=\frac{\sigma}{\epsilon_0}$ directed from positive plate to negative plate.

Potential difference between the plates of the capacitor is $V=Ed=\frac{\sigma d}{\epsilon_0}$ ( if the +ve plate is potential V then -ve plate is at potential 0)

a)

Initial kinetic energy of proton $K_i=\frac{1}{2}mu^2=0$

Initial potential energy of proton $P_i=eV=\frac{\sigma d e}{\epsilon_0}$

Final kinetic energy of proton $K_f=\frac{1}{2}mv^2$

Final potential energy of proton $P_f=0$

Conserving the energy of proton,

$K_i+P_i=K_f+P_f$

$0+\frac{\sigma d e}{\epsilon_0}=\frac{1}{2}mv^2+0$

$v=\sqrt{\frac{2\sigma d e}{\epsilon_0 m}}$

$v=\sqrt{\frac{2*1.8*10^{-7}*1.5*10^{-2}*1.6*10^{-19}}{ 8.85*10^{-12}*1.67*10^{-27}}}=2.42*10^5\,m/s$

------

b) Force on the proton in the electric field $E$ is  $F=eE$

$ma=eE\Rightarrow a=\frac{eE}{m}$

Using kinematics equation  $v^2-u^2=2ad \Rightarrow a=\frac{v^2-u^2}{2d}$

$\frac{v^2-u^2}{2d}=\frac{eE}{m}$

$v=\sqrt{\frac{2deE}{m}+u^2}=\sqrt{\frac{2de\sigma}{\epsilon_0m}+u^2}$

$v=\sqrt{\frac{2*1.5*10^{-2}*1.6*10^{-19}*1.8*10^{-7}}{8.85*10^{-12}*1.67*10^{-27}}+0^2}=2.42*10^5\,m/s$

------------------------------