Question

A particle of mass m and charge q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to momentary rest?

Answer 1

Answer:

-u^2m/2qe

Explanation:

f=qe

ma=qe

a=qe/m

we have

v^2=u^2+2as

s=u^2/2a

from equation 1

s=-u^2/2[qe/m]

s=-u^2m/2qe

Answer 2

Distance travelled will be $\frac {-u^2 m}{2qE}$

Explanation:

• Let us consider a particle of mass m and of charge q travelling against the uniform electric field E

then the force experience by the particle is

F = qE

ma = qE

$a = \frac {qE}{m}$ ""(1)

we have,

$v^2 = u^2 + 2as$

$0 = u^2 + 2as$

or $s = \frac {-u^2}{2a}$

From equation (1)-

s = $\frac {-u^2}{2\frac{qE}{m}}$

s = $\frac {-u^2 m}{2qE}$

therefore the distance travel by the particle before coming to momentary rest is given by $\frac {-u^2 m}{2qE}$ negative sign because the particle is retarding.