Question

One of the greatest physics experiments in history measured the charge-to-mass ratio of an electron, \(q / m\). If a uniform potential difference is created between two plates, atomized particles-each with an integral amount of charge- can be suspended in space. The assumption is that the particles of unknown mass, \(M\), contain a net number, \(n\), of electrons of mass \(m\) and charge \(q\). For a plate separation of \(d\), what is the potential difference necessary to suspend a particle of mass \(M\) containing \(n\) net electrons? What is the acceleration of the particle if the voltage is cut in half? What is the acceleration of the particle if the voltage is doubled?

Short answer

#Answer# When the voltage is cut in half: \(a_1 = \frac{\frac{1}{2}nq\frac{V}{d}}{M}\) When the voltage is doubled: \(a_2 = \frac{2nq\frac{V}{d}}{M}\)

Step-by-step solution

Analyze the forces acting on the particle

The particle of mass \(M\) containing \(n\) net electrons is suspended between two plates with a uniform potential difference. The force of gravity (weight \(W\)) acts downward on the particle, and the electric force \(F_e\) acts upward on the particle. When the particle is suspended, these forces are balanced.

Set up an equation that relates the electric force and gravity force

The electric force is calculated by \(F_e = nqE\), where \(E\) is the electric field between the plates. We know that \(E = \frac{V}{d}\), where \(V\) is the potential difference and \(d\) is the distance between the plates. Thus, we can write the electric force as \(F_e = nq\frac{V}{d}\). The weight of the particle is \(W = Mg\), where \(g\) is the acceleration due to gravity. Now we set the forces equal to find the necessary potential difference: \(Mg = nq\frac{V}{d}\).

Solve for the potential difference

To solve for the potential difference \(V\), we can rearrange the equation: \(V = \frac{Mgd}{nq}\).

Calculate the acceleration of the particle for different potential differences

First, let's find the initial electric force \(F_e\) when the potential difference is \(V\): \(F_e = nq\frac{V}{d}\). Now, the acceleration of the particle \(a\) can be calculated using Newton's second law: \(F_e = M a\). When the voltage is cut in half, the electric force is also cut in half. Thus, we can write the equation for the acceleration when the voltage is halved as: \(F_e' = \frac{1}{2}F_e = M a_1\) \(a_1 = \frac{F_e'}{M} = \frac{\frac{1}{2}F_e}{M} = \frac{\frac{1}{2}nq\frac{V}{d}}{M}\). When the voltage is doubled, the electric force is also doubled. Thus, we can write the equation for the acceleration when the voltage is doubled as: \(F_e'' = 2F_e = M a_2\) \(a_2 = \frac{F_e''}{M} = \frac{2F_e}{M} = \frac{2nq\frac{V}{d}}{M}\). These equations give the accelerations for the particle when the voltage is halved and doubled.