Question

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C. If the particles are free to move, what are their speeds (in m/s) after 47.6 ns?

Answer 1

Explanation:

Given that,

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C.

We need to find their speeds after 47.6 ns.

For electron,

The electric force is given by :

$F=qE\\\\F=1.6\times 10^{-19}\times 570\\\\=9.12\times 10^{-17}\ N$

Let a be the acceleration of the electron. So,

F = ma

m is mass of electron

$a=\dfrac{F}{m}\\\\a=\dfrac{9.12\times 10^{-17}}{9.1\times 10^{-31}}\\\\a=10^{14}\ m/s^2$

Let v be the final velocity of the electron. So,

v = u +at

u = 0 (at rest)

So,

$v=10^{14}\times 47.6\times 10^{-9}\\\\v=4.76\times 10^6\ m/s$

For proton,

Acceleration,

$a=\dfrac{9.12\times 10^{-17}}{1.67\times 10^{-27}}\\\\=5.46\times 10^{10}\ m/s^2$

Now final velocity of the proton is given by :

$v=5.46\times 10^{10}\times 47.6\times 10^{-9}\\\\v=2598.96\ m/s$

Hence, this is the required solution.