You are on the space station USS Oxford. And you are in the room where the is no

Question

You are on the space station USS Oxford. And you are in the room where the is normal Earth like gravity (10 m/s2 downward). You are to determine an E field in the room because when you 'drop' a charge, it has moved forward a distance by the time it reached the ground. Here is your data:

particle's mass is 125 micrograms. particle's charge is 30 nanocoulombs. intial velocity of the particle in x and y are both 0.0 meters per second (at rest). the particles initial height above ground was 1.4 meters. the particle moved forward 5 meters by the time it hit the ground. The E field is parallel to the floor.

WHat is the E field of the room and the final speed of the particle?

Explanation / Answer

Time taken by the mass to reach the ground = sqrt(2H/g)
= 0.53 sec

In this time it travelled 5 m in horizontal direction...
Let electric field be E.
Acceleration, a = F/m = qE/m
From equation of motion,
S = at2/2 = S = qEt2/2m

E = 2Sm/qt2

= (2*5*125*10^-6)/(.53*.53*30*10^-9)

= 37.08*10^3 N/C

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