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:

U_{0} + KE_{0} = U_{f} + KE_{f}

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

KE_{f} = (1/2)m(v/2)^{2}

KE_{0} = (1/2)mv^{2}

U_{f} = kq_{s}q_{e}/r

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

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 × 10^{7} 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?

Frequently Asked Questions

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

Our tutors have indicated that to solve this problem you will need to apply the Electric Potential Energy concept. You can view video lessons to learn Electric Potential Energy. Or if you need more Electric Potential Energy practice, you can also practice Electric Potential Energy practice problems.