Coulomb's Law (Electric Force) Video Lessons

Concept

# Problem: 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?

###### FREE Expert Solution

Newtons second law:

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

Coulomb's law:

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

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

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:

###### Problem Details

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?