The potential at a point `(x, y)` is expressed as `V = 4x + 3y`. A charge `1 mu C` and mass 1 mg is released from `(1, 1)` from rest. (a) Find the acceleration of the charge.
(b) Find the time when it crosses x axis.

Explanation

The potential at a given point can be differentiated to obtain the electric field. This electric field can then be used to find the acceleration of the charge.
Calculation:
`E_(x ) = (vartheta)/( vartheta x) and E_(y) = - (vartheta)/(varthetay)`
`overset(to) (E) = - 4 overset(^)(i) - 3 overset(^)(j)`
`overset(to) ( a) = (q overset(to) (E) )/ (m) = ((-4 overset(^)(i) - 4 overset(^)(j)) xx 10^(-6) )/(10^(-6))= -4 overset(^)(i) - overset(^)(j)`
Since the acceleration is constant, we can apply the equations of motion.
Calculation: According to the equation of motion:
`S= mu t + (1)/(2) "at"^2`.
When the charge crosses x axis, its y coordinate is 0. So, displacement in y direction is 0 - 1 = - 1 m. Hence,
`-1= (3)/(2) t^(2) rArr t = sqrt((2)/( 3)) S.`
Learn: The charged particle would move in a straight line.