Recent content by scarletx09

ok. the equilibrium position would occur when Fspring = Fefield --> -K(x-s0) = QE (x-s0) = (QE)/-k plugged that into my first diff. eq., the K canceled and i got: d^2t/dt^2 = (2QE)/m

Thanks so much for your guidance with part a! From my understanding of Part B: the first equilibrium position was at s0 which led to that diff eq. and now its asking for a different (smater i guess?) diff. eq. based on a new equilibrium position with the electric field "on" I am confused...

Using F=ma I got: Net force = Force of the electric field + Force of spring F = -QE - K(x-S0) = ma (x-s0) = the displaced length from equilibrium that resulted in: a = (-QE -Kx + K(s0))/m Does that seem correct? can i set up the equation of motion from this?

Homework Statement A mass m that has net electric charge Q is oscillating along the x-direction on one end of a spring (whose other end is anchored) of relaxed length s0. Suppose that someone then swirches on an electric field E  that is uniform in space, constant in time, and which points...

Homework Statement A thin, uniform 3.70- bar, 90.0 long, has very small 2.50- balls glued on at either end (the figure ). It is supported horizontally by a thin, horizontal, frictionless axle passing through its center and perpendicular to the bar. Suddenly the right-hand ball becomes...