Newtons second law:

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{m}}{\mathbf{a}}}$

Coulomb's law:

$\overline{)\begin{array}{rcl}{\mathit{F}}& {\mathbf{=}}& \frac{{\mathbf{kq}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\end{array}}$

m_{e} = 9.1 × 10^{-31} kg

k = 9.0 × 10^{9} N•m^{2}/C^{2}

q_{s} = -4.6 nC(10^{-9}C/1nC) = -4.6 × 10^{-9} C

q_{e} = -1.6 × 10^{-19} C

r = D/2 + L = (2.6 mm/2) + 0.37 mm = 1.67 mm(1m/1000mm) = 1.67 × 10^{-3} m

Combining Coulomb's law and Newton's second law:

A 2.6 mm -diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere from far away comes to within 0.37 mm of the surface of the target before being reflected.

What is the acceleration of the electron at its turning point?

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