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Sagot :
Answer: [tex]1.96\times 10^{8}\ m/s[/tex]
Explanation:
Given
Diameter of sphere is [tex]d=2.6\ mm\quad \quad[r=1.3 mm][/tex]
Charge on the sphere is [tex]Q=-4.5\ nC[/tex]
Nearest distance electron can reach to sphere is [tex]d=0.37\ mm[/tex]
Here, kinetic energy of electron is converted into electrostatic energy between the two i.e.
[tex]\Rightarrow \dfrac{1}{2}mv^2=\dfrac{kQq}{d}\\\\\Rightarrow \dfrac{1}{2}\times 9.1\times 10^{-31}\times v^2=\dfrac{9\times 10^9\times 4.5\times 10^{-9}\times 1.6\times 10^{-19}}{(3.7\times 10^{-4})}\\\\\Rightarrow v^2=3.8491\times 10^{16}\\\Rightarrow v=1.96\times 10^{8}\ m/s[/tex]
Thus, the initial speed of electron is [tex]1.96\times 10^{8}\ m/s[/tex].
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