There is a uniform electric field of strength `10^(3)V//m` along `y`-axis. A body of mass `1g` and charge `10^(-6)C` is projected into the field from origin along the positive `x`-axis with a velocity `10 m//s`. Its speed in `m//s` after `10 s` is (Neglect gravitation)

Options

A. 10
B. `5sqrt2`
C. `10sqrt2`
D. 20

Answer

C

Explanation

The charged particle will be accelectred along the y-axis, as the electric field acts along y-axis.
`therefore a_(y) =("Force due to electric field ")/("Mass")=(qE)/(m)`
`=(10^(-6) xx10^(3))/(10^(-3))=1 m //s^(2) " "[because E = F /q]`
`therefore` Vertical velocity of the particle after 10 second is
`v_(y) =a_(y)t=1 xx10=10m//s`
and the horizontal velocity `v_(x)` remains constant and `v_(x) =10m//s`
`therefore` Resultant velocity `= sqrt(v _(x)^(2) +v_(y)^(2))=sqrt(10^(2) +10^(2))`
`therefore v = 10sqrt2 m//s`