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# Problem: A 2.6 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away with an initial velocity of  9.34 × 107 m/s. It comes to rest within 0.37 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?

###### FREE Expert Solution

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{E}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}}^{{\mathbf{2}}}}$

Electric potential energy:

$\overline{)\begin{array}{rcl}{\mathbf{U}}& {\mathbf{=}}& \frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}\end{array}}$

Conservation of energy:

U0 + KE0 = Uf + KEf

U0 = 0 J. The separation, r, is very large such that the potential energy is zero.

KEf = (1/2)m(v/2)2

KE0 = (1/2)mv2

Uf = kqsqe/r

me = 9.1 × 10-31 kg

k = 9.0 × 109 N•m2/C2

qs = -4.6 nC(10-9C/1nC) = -4.6 × 10-9 C

qe = -1.6 × 10-19 C

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###### Problem Details

A 2.6 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away with an initial velocity of  9.34 × 107 m/s. It comes to rest within 0.37 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?

What scientific concept do you need to know in order to solve this problem?

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