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

Answer:

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

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

⇒$\frac{m}{2}(v^2-u^2 ) = qE(l_i - l_f)$

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