Electric potential and charged particles in fields - TCE Physics (Tasmania)

What this dot point is asking

This dot point connects fields to energy and motion: how much energy a charge gains crossing a potential difference, and how a charged particle moves once inside a field.

Work and potential difference

When a charge $q$ moves through a potential difference $V$, the work done by the field is:

If the charge starts at rest and is free to move, all this work becomes kinetic energy:

This is how electron guns and particle accelerators give charges high speeds: a known voltage accelerates them to a calculable speed. Inside a uniform field the force is constant, so you can also find the work as $W = qEd$ over a distance $d$ along the field, and $V = Ed$ links the two views.

Electric potential $V$ at a point is the potential energy per unit charge placed there, measured in volts ($\text{J C}^{-1}$). The potential difference between two points is what does work on a charge moving between them, which is why only differences in potential, not absolute values, appear in calculations. In a uniform field the potential drops steadily from the positive plate to the negative plate, and the field points from high potential to low potential.

Charges accelerated along the field

A charge released in a uniform field accelerates along the field lines (positive charges follow the field, negative charges go against it). The acceleration is constant:

so the constant-acceleration equations of motion apply directly. This is the simplest case: straight-line acceleration, like free fall but driven by the electric force instead of gravity. Combined with $V = Ed$, you can swap freely between an energy view ($qV = \tfrac12 m v^2$) and a kinematic view ($v^2 = u^2 + 2as$) of the same accelerated charge.

Charges fired across the field

If a charge enters a uniform field moving perpendicular to it, the motion splits into two independent parts, just like a projectile:

• Along the original direction the velocity is constant (no force that way).
• Across the field the charge accelerates uniformly under $F = qE$.

The result is a parabolic path. This is exactly how the old cathode ray tube deflected electrons to paint a picture, and it is a favourite exam scenario because it combines fields with projectile analysis. The vertical deflection while the charge is between plates of length $\ell$ is $y = \tfrac12 \dfrac{qE}{m}\left(\dfrac{\ell}{u}\right)^2$, found by using the time of flight $t = \dfrac{\ell}{u}$ from the constant horizontal motion.

In the exam, for a charge accelerated through a voltage use the energy method $qV = \tfrac12 mv^2$. For a charge fired across a field, separate the motion into constant-velocity and constant-acceleration components and treat it as a projectile with acceleration $\dfrac{qE}{m}$. Keep the geometry of the plates clearly in mind so you use the right length for the time of flight.