A uniform field of `2 kN//C` is the x direction. A point charge `= 3 mu C` initially at rest at the origin is released. What is K.E. of this charge at x = 4m ? Also, calculate `V (4m) - V (0)`.

Answer

24xx10^(17) Vm^(-1) ; 4.5xx10^(21) Vm^(-1)`

Explanation

From `E = - (dV)/(dr)`,
`dV = -E (dr) = -2xx10^(3) (4) = -8xx10^(3) V`
Gain in K.E. = Loss in P.E.
`= q (dV) = 3xx10^(3) (8xx10^(3)) = 24xx10^(-3)J`