Answer to Question #320174 in Electricity and Magnetism for Ron

Question

A charge of 4.50x10^-8C is placed in a uniform electric field that is directed vertically upward with a magnitude of 5.00x10^4N/C. What work is done by the electrical force when the charge moves 0.450m to the right? 0.800m downward? 2.60m at an angle of 45° from the horizontal?

Answer 1

(1)

Work

$W=Fdcos\theta$

$$W=qEdcos\theta\\W=4.50\times10^{-8}\times5\times10^4\times0.450\times cos 90=0J$$

$$W=qEdcos\theta\\W=4.50\times10^{-8}\times5\times10^4\times0.800\times cos 0=1.8\times10^{-3}J$$

$$W=qEdcos\theta\\W=4.50\times10^{-8}\times5\times10^4\times2.60\times cos 45°=4.13\times10^{-3}J$$