Thomson used a beam of negatively charged particles.
Answer:
I will answer this in English, we can translate it to:
Why if you charge a mate by an amount of time you are not doing work?
This happens because work is defined as the displacement done by a force:
W = d*F
where W is work, d is the distance, and F is the force.
This means that the amount of time that you are charging your mate does not affect the mechanical work, the only time that you are doing work is when you are lifting him.
In order for particles to perform a simple harmonic motion, we must follow the law of force of the form F = -kx, where x is the displacement of the object from the equilibrium position and k is the spring constant. The
force shown in <span>F = -kx is always the restoring force in the sense
that the particles are pulled towards the equilibrium position.
The
repulsive force felt when the charge q1 is pushed into another charge
q2 of the same polarity is given by Coulomb's law
F = </span><span>k *q1* q2 / r^2.
</span>It is clear that Coulomb's law is an inverse-square relationship. It does not have the same mathematical form as the equation <span><span>F = -kx.</span> Thus,
charged particles pushed towards another fixed charged particle of
the same fixed polarity do not show a simple harmonic motion when
released. Coulomb's law does not describe restoring force. When q1 is released, it just fly away from q2 and never returns.</span>
It does work or increases thermal energy
