Question

Part B At what distance from the surface of the sphere is the electron's speed half of its initial value? Express your answer using two significant figures. Submit My Answers Give Up Part C What is the acceleration of the electron at its turning point? Express your answer using two significant figures. a= m/s Submit My Answers Give Up

Answer 1

b) At the point of maximum closest approach, electron's kinetic energy will be equal to the electrostatic potential energy 9.1x10-31 kg x(1.3x10-3 m +0.34x10-3 m -9.21x10 ms1 Let x be the distance from the surface of the sphere at which electron will have half the initial speed 4.4×10-9 C×1.6× 10-19 C 1.3×10-3 m + x 9×109 Nm"C-2 × 4.4×10-9 ×1.6×10-19 C 2 1.3x10-3 m +x 8×9×109 Nm-C-2×4.4×10-9×1.6×10-19 C 1.3x10-, n-x 3x9.1x10- kg x(9.21x10 ms1 0.0202 nm Hence, distance from the surface of the sphere where the electron's speed become half is 0.02 m

c) Force acting on the electron at the turning point is given by, 1 qe (1.3x101m +o.34xl0'm) -2.36x10-12 N 2.36x10-12 N 9.1x10-1 kg 2.59 x105 ms2 Hence, acceleration of the electron at the turning point is 2.6x10's ms