Question

An electron entering the field normally with a velocity 4*10 ^7 m/stravels a distance of 0.10 m in an electric field of 3200v/m what is the deviation from its path

Answer 1

Answer:

y = 0.5*(eE/m)*t^2

put the value of charge mass and given electric field inensity

and get the reln. in y and t.

 

Calculate t from:

x = v*t

0.10 = 4*10^7*t

 

and put it in y eqn. and the deviation value .i.e. y.

Explanation:

Answer 2

The deviation from its path is $1.75\times 10^{-3}\ V$.

Explanation:

It is given that,

Velocity of the electron, $v=4\times 10^7\ m/s$

Distance covered, d = 0.1 m

Electric field, E = 3200 V/m

Time taken by the electron to travel a distance of 0.1 m is given by :

$t=\dfrac{d}{v}$

$t=\dfrac{0.1\ m}{4\times 10^7\ m/s}$

$t=2.5\times 10^{-9}\ s$

If a is the acceleration of the electron. In the electric field acceleration of the electron is given by :

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 3200}{9.1\times 10^{-31}}$

$a=5.62\times 10^{14}\ m/s^2$

Let D is the deviation from its path. It can be calculated using second equation of motion as :

$D=\dfrac{at^2}{2}$

$D=\dfrac{5.62\times 10^{14}\times (2.5\times 10^{-9})^2}{2}$

$D=1.75\times 10^{-3}\ V$

So, the deviation from its path is $1.75\times 10^{-3}\ V$. Hence, this is the required solution.

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