We know that electric field (E ) at any point in space can be calculated using the relation
`vecE = - (deltaV)/(deltax)hati - (deltaV)/(deltay)hatj - (deltaV)/(deltaz)hatk`

if we know the variation of potential (V) at that point. Now let the electric potential in volt along the x-axis vary as `V = 2x^2`, where x is in meter. Its variation is as shown in figure
A charge particle of mass 10 mg and charge `2.5 muC` is released from rest at `x = 2 m`. Find its velocity when it crosses origin.

Options

A. `0.5ms^(-1)`
B. `1 ms^(-1)`
C. `2ms^(-1)`
D. `4 ms^(-1)`

Answer

C

Explanation

`E=-(dV)/(dx)=-4x`
`F=qE=2.5xx10^(-6)(-4x)=-10^(-5)x`
`W=underset(2)overset(0)int Fdx=-10^(-5)underset(2)overset(0)intxdx`
`(1)/(2)mv^(2)=10^(-5)[(x^(2))/(2)]_(0)^(2)` or `V=2ms^(-1)`