Question

A charged particle is accelerated from rest through a potential difference \(\Delta V\). The particle then passes straight through a velocity selector (field magnitudes \(E\) and \(B\) ). Derive an expression for the charge-to-mass ratio \((q / m)\) of the particle in terms of \(\Delta V, E,\) and \(B\)

Short answer

Answer: The expression for the charge-to-mass ratio (q/m) is given by: $$\frac{q}{m} = \frac{1}{\Delta V} \cdot \frac{1}{2} \left(\frac{E}{B}\right)^2$$

Step-by-step solution

Kinetic energy gained by the particle

The charged particle is accelerated from rest through a potential difference \(\Delta V\). We can use the formula for energy conservation to say that the electric potential energy is converted into kinetic energy $$q \Delta V = \frac{1}{2}mv^2$$ Where \(q\) is the charge of the particle, \(m\) is the mass of the particle, and \(v\) is the final particle velocity.

Use the velocity selector condition

The velocity selector is designed such that the electric force acting on the charged particle is balanced by the magnetic force. This will allow only particles with a specific velocity \(v\) to pass straight through the selector. We can write the balance of the forces as $$ F_{electric} = F_{magnetic} $$ where \(F_{electric} = qE\) (electric force) and \(F_{magnetic} = qvB\) (magnetic force), so $$ q E = q v B $$ Now we can find the particle's velocity \(v\) in terms of the charge \(q\), field magnitudes \(E\) and \(B\), as $$v = \frac{E}{B}$$

Substitute the velocity expression in the energy conservation equation

Now that we have found an expression for the particle's velocity, we can substitute it into the energy conservation equation $$q \Delta V = \frac{1}{2}m \left(\frac{E}{B}\right)^2$$

Solve for the charge-to-mass ratio \((q/m)\)

We can now solve this equation for the charge-to-mass ratio \((q/m)\) by dividing through by \(m\) and moving the potential difference term \(\Delta V\) to the other side. $$\frac{q}{m} = \frac{1}{\Delta V} \cdot \frac{1}{2} \left(\frac{E}{B}\right)^2$$ Hence, we have derived an expression for the charge-to-mass ratio \((q/m)\) in terms of the potential difference \(\Delta V\), and field magnitudes \(E\) and \(B\) as $$\frac{q}{m} = \frac{1}{\Delta V} \cdot \frac{1}{2} \left(\frac{E}{B}\right)^2$$