Auniform electric field of magnitude 300 v/m is directed in the negative x-direction. a 20 µc, 10 ? g particle moves from the origin to the point (x, y) = (20 cm, 50 cm).
(a) through what electric potential difference did the particle move?

(b) what was the work done on this particle?

(c) if the particle's speed was 600 m/s at the origin, what is it's speed at the point (x, y) = (20 cm, 50 cm)?
/s

Distance covered = speed x timethe distance traveled going against the stream = distance traveled with the stream.therefore it is the speed and the time that are the variables.as he travels the same distance both ways, speed x time is the same for each way.going upstream his speed is 4 kph; going downstream his speed is 6 kph.therefore we have two simultaneous equations: time upstream x 4 = time downstream x 6time upstream + time downstream = 60 minutestherefore we have the equation: time upstream = 6/10 hour, time downstream therefore = 4/10 hour.distance traveled at 4km/hr in 6/10 hour = 24/10 = 2.4km.check: distance traveled at 6km/hr in 4/10 hour = 24/10 = 2.4km.

Answervery cold;