Question

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 564 N/C. If the particles are free to move, what are their speeds (in m/s) after 49.6 ns?

Answer 1

For electron

`V=4.9* 10^6\ m/s`

For proton

`V=2680.18\ m/s`

Step-by-step explanation:

Given that

E= 564 N/C

Charge on electron

`q=-1.6* 10^(-19)\ C`

Mass of electron

`m=9.10* 10^(-34)\ kg`

Charge on proton

`q=1.6* 10^(-19)\ C`

Mass of proton

`m=1.67* 10^(-27)\ kg`

We know that

F = E q

F= m a

For electron:

F= E q

m a = E q

`9.10* 10^(-34)\ a=564* 1.6* 10^(-19)`

`a=9.9* 10^(16)\ m/s^2`

V= U + a t

`V=9.9* 10^(16)* 49.6* 10^(-9)\ m/s`

`V=4.9* 10^6\ m/s`

For proton:

F= E q

m a = E q

`1.67* 10^(-27)\ a=564* 1.6* 10^(-19)`

`a=5.4* 10^(10)\ m/s^2`

V= U + a t

`V=5.4* 10^(10)* 49.6* 10^(-9)\ m/s`

`V=2680.18\ m/s`