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Sagot :
Explanation:
Given that,
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C.
We need to find their speeds after 47.6 ns.
For electron,
The electric force is given by :
[tex]F=qE\\\\F=1.6\times 10^{-19}\times 570\\\\=9.12\times 10^{-17}\ N[/tex]
Let a be the acceleration of the electron. So,
F = ma
m is mass of electron
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{9.12\times 10^{-17}}{9.1\times 10^{-31}}\\\\a=10^{14}\ m/s^2[/tex]
Let v be the final velocity of the electron. So,
v = u +at
u = 0 (at rest)
So,
[tex]v=10^{14}\times 47.6\times 10^{-9}\\\\v=4.76\times 10^6\ m/s[/tex]
For proton,
Acceleration,
[tex]a=\dfrac{9.12\times 10^{-17}}{1.67\times 10^{-27}}\\\\=5.46\times 10^{10}\ m/s^2[/tex]
Now final velocity of the proton is given by :
[tex]v=5.46\times 10^{10}\times 47.6\times 10^{-9}\\\\v=2598.96\ m/s[/tex]
Hence, this is the required solution.
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