Question

An electron and a proton are each placed at rest in a uniform electric field of magnitude 554 N/C. Calculate the speed of each particle 52.0 ns after being released. electron m/s proton m/s Need Help?

Answer 1

Explanation:

It is given that,

Electric field, E = 554 N/C

Time, $t=52\ ns=52\times 10^{-9}\ s$

Electric force, F = qE

For both electron and proton, $F=1.6\times 10^{-19}\ C\times 554\ N/C$

$F=8.86\times 10^{-17}\ N$

For electron, $F=m_ea_e$

$a_e=\dfrac{F}{m_e}$

$a_e=\dfrac{8.86\times 10^{-17}\ N}{9.1\times 10^{-31}\ kg}$

$a_e=9.73\times 10^{13}\ m/s^2$

Using first equation of motion as :

$v=u+at$

u = 0

$v=9.73\times 10^{13}\ m/s^2\times 52\times 10^{-9}\ s$

v = 5059600 m/s

or

v = 5.05 × 10⁶ m/s

For proton :

$F=m_pa_p$

$a_p=\dfrac{F}{m_e}$

$a_p=\dfrac{8.86\times 10^{-17}\ N}{1.67\times 10^{-27}\ kg}$

$a_p=5.3\times 10^{10}\ m/s^2$

Using first equation of motion as :

$v=u+at$

u = 0

$v=5.3\times 10^{10}\ m/s^2\times 52\times 10^{-9}\ s$

v = 2756 m/s

Hence, this is the required solution.