Question

An electron with a speed of 3.00 106 m/s moves into a uniform electric field of magnitude 1.00 103 N/C. The field lines are parallel to the electron's velocity and pointing in the same direction as the velocity. How far does the electron travel before it is brought to rest?

Answer 1

0.0256 m

Step-by-step explanation:

initial speed of electron, u = 3 x 10^6 m/s

Final speed of electron, v = 0 m/s

Electric field strength, E = 1 x 10^3 N/C

Charge on electron, q = 1.6 x 10^-19 C

mass of electron, m = 9.1 x 10^-31 kg

The electron experiences an electric force due to electric field which is given by

F = q E

Where, f be the electric force, and E be the strength of electric field.

Let a be the acceleration of electron

So,
`a= (F)/(m)=(qE)/(m)`

where, m is the mass of electron

By substituting the values, we get

`a=(-1.6*10^(-19)*10^(3))/(9.1*10^(-31))=-1.758 * 10^(14)m/s^2`

Let the distance traveled by the electron before coming to rest is s.

Use third equation of motion

`v^(2)=u^(2)+2as`

`0^(2)=\left (3* 10^6  \right )^(2)-2* 1.758 *10^(14)* s`

s = 0.0256 m