A 2.3 mm -diameter sphere is charged to -4.4 nC . An electron fired directly at

Question

A 2.3 mm -diameter sphere is charged to -4.4 nC . An electron fired directly at the sphere from far away comes to within 0.35 mm of the surface of the target before being reflected. Part A What was the electron's initial speed? Express your answer using two significant figures. 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. Part C What is the acceleration of the electron at its turning point? Express your answer using two significant figures.

Explanation / Answer

Here,

radius of sphere , r = 2.3/2 mm

r = 1.15 mm

Q = -4.4 nC

let the inital speed of electron is u m/s

Using conservation of energy

k * Q * q/d = 0.5 * m * u^2

9*10^9 * 4.4 *10^-9 * 1.602 *10^-19/(1.15 *10^-3 + 0.35 *10^-3) = 0.5 * 9.109 *10^-31 * u^2

solvng for u

u = 9.637*10^7 m/s

the initial speed of electron is 9.637*10^7 m/s

part b)

for one half of the speed ,

let the distance is d

Using conservation of energy

k * Q * q/(r + d) = 0.5 * m * (u/2)^2

9*10^9 * 4.4 *10^-9 * 1.602 *10^-19/(1.15 *10^-3 + d) = 0.5 * 9.109 *10^-31 * (9.637*10^7 /2)^2

solving for d

d = 0.00485 m = 4.85 mm

the distance from the surface will be 4.85 mm

part c)

let the acceleration is a

a = force/mass

a = 9*10^9 * 4.4 *10^-9 * 1.602 *10^-19/((1.15 *10^-3 + 0.35 *10^-3)^2 * 9.11 *10^-31 )

a = 3.095 *10^18 m/s^2

the acceleration of electron at the turn about point is 3.095 *10^18 m/s^2

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