Respuesta :
Answer:
Electron will travel to a distance of 0.189 m before its velocity is 0
Explanation:
We have initial speed of the electron [tex]u=27.5\times 10^6m/sec[/tex]
Electric field [tex]E=11.4\times 10^3N/C[/tex]
Charge on electron will be [tex]q=1.6\times 10^{-19}C[/tex]
Mass of electron [tex]m=9.1\times 10^{-31}kg[/tex]
We know that Columbus force is equal to F = qE and according to newtons law force is equal to F = ma , here m is mass and a is acceleration
These two forces will be equal
So ma = qE
[tex]a=\frac{qE}{m}=\frac{1.6\times 10^{-19}\times 11.4\times 10^3}{9.1\times 10^{-31}}=2\times 10^{15}m/sec^2[/tex]
According to third equation of motion
[tex]v^2=u^2+2as[/tex]
So [tex]0^2=(27.5\times 10^{6})^2+2\times 2\times 10^{15}\times s[/tex]
s = 0.189 m
So electron will travel to a distance of 0.189 m before its velocity is 0
The electron will travel a distance of 0.189 m.
Given that the initial speed of the electron is [tex]v_0=27.5\times10^6[/tex] m/s.
Electric field E = 11.4 × 10³ N/C
Electrostatic Force:
The force due to the electric field on the electron is given by:
ma = qE
where m = 9.1 × 10⁻³¹ kg is the mass of the electron
q = 1.6 × 10⁻¹⁹ C is the charge of the electron, and
a is the acceleration.
So,
[tex]a=\frac{qE}{m}\\\\a=\frac{1.6\times10^{-19}\times11.4\times10^3}{9.1\times10^{-31}}\\\\a=2\times10^{15}m/s^2[/tex]
Now the final speed must be v = 0
Applying third equation of motion:
[tex]v^2=v_0^2-2as\\\\0=(27.5\times10^6)^2-2\times(2\times10^{15})\times s[/tex]
s = 0.189 m
Learn more about electrostatic force:
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