A 2.5 mm -diameter sphere is charged to -4.4 nC. An electron fired directly at t

Question

A 2.5 mm -diameter sphere is charged to -4.4 nC. An electron fired directly at the sphere from far away comes to within 0.39 mm of the surface of the target before being reflected.

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.

*The electron's initial speed is 9.2 x 10^7 m/s

Somebody had given me this but I'm having trouble solving through it and not sure if its even correct:

(1/2)mv2 =(1/2)mv2+kqe/(r+x) here v =v/2   and here r =1.25*10-3m

Now find x by substituting the all the values

Explanation / Answer

Apply law of conservation of energy

0.5*m*u^2 = -k*q*e/(x+r1)

0.5*m*(9.2*9.2)*10^14 = 9*10^9*4.4*10^-9*1.6*10^-19/((0.39 + 1.25)*10^-3) = 38.63*10^-16

m = 0.912*10^-30 kg

thena agian apply

0.5*m*u^2 = 0.5*m*(u/2)^2 - k*q*e/(x+r)

k*q*e/(x+r) = (-3/4)*0.5*m*u^2 = (-3/4)*0.5*0.912*10^-30*9.2*9.2*10^14 = 28.97*10^-16

x+r = (-9*10^9*1.6*10^-19*4.4*10^-9)/(28.97*10^-16)

x+r = 2.18*10^-3 m

x = (2.18 - 1.25)mm = 0.93 mm

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