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Consider an electron on the surface of a uniformly charged sphere of radius 1.2 cm and total charge 1.4 10-15 C. What is the “escape speed”
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Answer:
Explanation:
electric potential of electron
= k Qq / r
= – 9 x 10⁹ x 1.4 x 10⁻¹⁵ x 1.6 x 10⁻¹⁹ / ( 1.2 x 10⁻² )
= – 16.8 x 10⁻²³ J
If v be the escape velocity
1/2 m v² = potential energy of electron
= 1/2 x 9.1 x 10⁻³¹ x v² = 16.8 x 10⁻²³
v² = 3.69 x 10⁸
v = 1.92 x 10⁴ m /s