Question

a proton travelling along is x-axis is lowed by a niform electric field E. at x = 20.0 cm, the proton has a speed of 3.5x10^6 m/s and at 80.0 cm the speed is zero. Determine the magnitude and direction of e,

Answer 1

Magnitude of electric field is 1.06 x
`10^5` V/m along negative X-direction

Step-by-step explanation:

Given: initial velocity of proton = u = 3.5 x
`10^6` m/s

final velocity of proton = v = 0 m/s

initial point
`l_i` = 0.2 m and final point is
`l_f` = 0.8 m

According to conservation of energy:

change in in kinetic energy = change in potential energy of proton

⇒
`(m)/(2)(v^2-u^2 ) = qE(l_i - l_f)`

where q and m is the charge and mass of proton E is the electric field ,
`l_i` and
`l_f` is the initial and final position of proton

on substituting the respected values we get,

1.023 x
`10^-^1^4` = 9.6 x
`10^-^2^0` x E

⇒ E = 1.06 x
`10^5` V/m

external force is opposite to the motion as velocity of proton decreases with distance.

Therefore, magnitude of electric field is 1.06 x
`10^5` V/m along negative X-direction