Question

A particle has charge -3.00 nC. (a) Find the magnitude

nd direction of the electric field due to this particle at a point

.250 m directly above it. (b) At what distance from this particle

pes its electric field have a magnitude of 12.0 N/C?

Answer 1

Explanations & Calculations

a)

• Apply Coulomb's law to the particle,

$\qquad\qquad \begin{aligned} \small E&=\small k.\frac{Q}{r^2}\\ &=\small (9\times10^9\,Nm^2C^{-2}).\frac{3.00\,\times10^{-9}\,C}{(0.250\,m)^2}\\ &=\small 432\,NC^{-1} \end{aligned}$

• Directed directly down from that point towards the charge. This is because the charge is negatively charged and the negative charges have all the field lines directed towards them.

b)

• Apply the same formula

$\qquad\qquad \begin{aligned} \small 12\,NC^{-1}&=\small (9\times10^9\,Nm^2C^{-2}).\frac{3.00\times10^{-9}\,C}{r^2}\\ \small r&=\small 1.500\,m \end{aligned}$