point particle with charge Q and mass M is subjected to a ti

point particle with charge Q and mass M is subjected to a time-dependent
electric field which points in the ˆx direction, with
Ex = E0 sin(Bt)
where E0 and B are constants. Assume the particle starts at x0 and has velocity v0 in the
positive ˆx direction at time t = 0. The motion is in one dimension since y = z = 0 at all
times. Find x(t).

Solution

We need to consider the fact that the charge under the influence of the electric field would suffer a net force along the +X axis and hence would undergo acceleration in that direction.

For t = 0, we have velocity = Vo

Now, for any time t, the electric field would be given by EoSin(Bt)

Hence, the force suffered by the charge Q, would be QEoSin(Bt),

that is, the acceleration = (QEo/M) Sin(Bt)

We also know that, dv/dt = acceleration

That is, dv/dt = (QEo/M) Sin(Bt)

or, dv = (QEo/M) Sin(Bt) dt

We integrate the both sides above, from t =0 to t and velocity from Vo to V

V - Vo = (QEo/M) (1 - Cos(Bt))

or V(t) = Vo + (QEo/M) (1 - Cos(Bt))

Further, dx/dt = V(t) = Vo + (QEo/M) (1 - Cos(Bt))

or dx = [Vo + (QEo/M) (1 - Cos(Bt))] dt

Integrating the both sides we obtain:

x(t) = Vot + (QEo/M)(t - Sin(Bt))) which is the required relation.