Charged particles and simple harmonic motion.

A particle of mass "m", charge "-q" lies in the middle of the distance two fixed particles of charge "+q". The distance between the two positive particles is denoted by L. The negative particle is pulled upwards in the "y" axis and then released. Show that in the condition in which L is much larger than the oscillation amplitude of the negative charge, the net force on the negative particle is given by:

Fnet = (-1/2πε0) * (q²y(t)/(L/2)³).

Use the following relation if necessary: (1+x)^n = 1 + nx.

Question

Charged particles and simple harmonic motion.

A particle of mass "m", charge "-q" lies in the middle of the distance two fixed particles of charge "+q". The distance between the two positive particles is denoted by L. The negative particle is pulled upwards in the "y" axis and then released. Show that in the condition in which L is much larger than the oscillation amplitude of the negative charge, the net force on the negative particle is given by:

Fnet = (-1/2πε0) * (q²y(t)/(L/2)³).

Use the following relation if necessary: (1+x)^n >= 1 + nx.

chillout · Answer

Ive got it! Will post the solution then close it.

chillout · Answer

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