QCAA Physics Electromagnetism

$F = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r^2}$
$0.025 = \frac{9 \times 10^9 \times Q^2}{0.30^2}$
$Q = \sqrt{\frac{0.025 \times 0.30^2}{9 \times 10^9}}$
$= 5.0 \times 10^{-7}$
Charge = $5.0 \times 10^{-7} \text{ C}$

$F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2}$
$r^2 = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{F}$
$r = \sqrt{\frac{1}{4\pi\varepsilon_0} \frac{Qq}{F}}$
$= \sqrt{9 \times 10^9 \times \frac{2.8 \times 10^{-7} \times 3.2 \times 10^{-7}}{0.52}}$
Distance = 0.039 m (to 2 significant figures)

The change in potential energy of an electric charge moving through an electric field is equivalent to the work done on the charge.
$V = \frac{\Delta U}{q}$
$\Delta U = Vq$
$= 240 \times 3.2 \times 10^{-19}$
$= 7.68 \times 10^{-17} J = W$

The work done on an object is equal to the change in kinetic energy.
$E_k = \frac{1}{2}mv^2$
$7.68 \times 10^{-17} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2$
$v^2 = \frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}$
$v = \sqrt{\frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}}$
Velocity = $1.5 \times 10^5 \text{ m s}^{-1}$ (to 2 significant figures)