In a Millikan oil-drop experiment, a uniform electric field of 5.71 x 10^5 N/C is maintained in the region between two plates separated by 6.49 cm. Find the potential difference (in V) between the plates.

37057.9V

Explanation:

Electric potential is defined as the work done in moving a unit positive charge from infinity to a point.

Electric potential (E) = Potential Difference (V)/distance between plates(d)

Given; electric field of 5.71 x 10^5 N/C; distance between plates =6.49cm = 0.0649m

Since E = V/d

V = Ed

V = 5.71×10^5×0.0649

V = 37057.9Volts

The potential difference (in V) between the plates is 37057.9V

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s