Question

An electron and a proton are each placed at rest in a uniform electric field of magnitude 560 N/C. Calculate the speed of each particle 46.0 ns after being released. electron 4.5e^-6 Incorrect.

Answer 1

Answer:

The speed of electron is $v=4.52\times 10^6\ m/s$ and the speed of proton is 2468.02 m/s.

Explanation:

Given that,

Electric field, E = 560 N/C

To find,

The speed of each particle  (electrons and proton) 46.0 ns after being released.

Solution,

For electron,

The electric force is given by :

$F=qE$

$F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N$

Let v is the speed of electron. It can be calculated using first equation of motion as :

$v=u+at$

u = 0 (at rest)

$v=\dfrac{F}{m}t$

$v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}$

$v=4.52\times 10^6\ m/s$

For proton,

The electric force is given by :

$F=qE$

$F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N$

Let v is the speed of electron. It can be calculated using first equation of motion as :

$v=u+at$

u = 0 (at rest)

$v=\dfrac{F}{m}t$

$v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}$

$v=2468.02\ m/s$

So, the speed of electron is $v=4.52\times 10^6\ m/s$ and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.