The escape speed of an electron launched from | vedclass

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(A) The escape speed $v$ is determined by the conservation of energy,where the initial kinetic energy equals the potential energy at the surface: $\fr

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The escape speed $v$ is determined by the conservation of energy,where the initial kinetic energy equals the potential energy at the surface: $\frac{1}{2}mv^2 = \frac{1}{4\pi \epsilon_0} \frac{Qe}{R}$.Solving for $v$: $v = \sqrt{\frac{2}{4\pi \epsilon_0} \frac{Qe}{mR}}$.Given: $Q = 10 	imes 10^{-9} \ C$,$R = 0.5 	imes 10^{-2} \ m$,$e = 1.6 	imes 10^{-19} \ C$,$m = 9.1 	imes 10^{-31} \ kg$,and $\frac{1}{4\pi \epsilon_0} = 9 	imes 10^9 \ N \cdot m^2/C^2$.Substituting the values: $v = \sqrt{2 	imes (9 	imes 10^9) 	imes \frac{(10 	imes 10^{-9}) 	imes (1.6 	imes 10^{-19})}{9.1 	imes 10^{-31} 	imes 0.5 	imes 10^{-2}}}$.$v = \sqrt{18 	imes 10^9 	imes \frac{16 	imes 10^{-28}}{4.55 	imes 10^{-33}}} = \sqrt{63.29 	imes 10^{14}} \approx 7.95 	imes 10^7 \ m/s$.Rounding to the nearest integer,$x \approx 8$.

The escape speed of an electron launched from the surface of a glass sphere of diameter $1 \ cm$ that has been charged to $10 \ nC$ is $x \times 10^7 \ m/s$. The value of $x$ is:

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