一様電気場

The oil drop that hung in mid-air

• In 1909 Robert Millikan sprayed tiny oil drops between two horizontal charged plates and adjusted the voltage until a drop hung perfectly still.
• Still meant the electric force exactly balanced the weight, which turned the drop into a scale for its own charge. Millikan measured hundreds, and every charge was a whole-number multiple of one tiny value.
• That value was $e$. A uniform field between two plates was enough to prove that charge comes in indivisible lumps.
• This lesson is the uniform field 匀强场 between parallel plates, what it does to a charged particle, and the two situations the exam sets: the balanced drop and the deflected beam.

Strength of a uniform field

• Between two parallel plates a distance $d$ apart with potential difference 电势差 $V$ between them, the field is uniform apart from edge effects:

• It points from the higher-potential plate to the lower one. The unit $\text{V/m}$ comes straight from this equation and equals $\text{N/C}$.
• Only two things change the field: increase the p.d., or move the plates closer. A resistor in series with the supply changes nothing, since no current flows once the plates are charged, so there is no p.d. across it.

Evenly spaced parallel lines: the same strength everywhere between the plates

A charged particle in the field

• The force $F = qE$ is the same everywhere, so the acceleration $a = qE/m$ is constant. This is exactly a mass in a uniform gravitational field, with $qE$ in place of $mg$.
• Released at rest, the particle accelerates in a straight line and gains kinetic energy. The energy method is usually quickest: $E_{\text{k}} = qV$, so $v = \sqrt{2qV/m}$, with no kinematics at all.
• Entering at right angles, it keeps its constant sideways speed while accelerating across the field, so it follows a parabola 抛物线, exactly like a projectile 抛体.

Constant velocity one way, constant acceleration the other

Worked example: an electron crossing the gap

Worked example: the balanced oil drop

Two ways to change a uniform field

• Between parallel plates the field is $E = V/d$, so the field depends on both the p.d. and the separation. What happens when you move the plates apart depends on what is held fixed, and the exam asks both versions.
• Connected to a supply, so $V$ is fixed: doubling $d$ halves $E$. The charge on the plates falls to match.
• Isolated after charging, so the charge is fixed: the field between the plates is unchanged, and it is $V$ that doubles instead.
• The second case surprises people every year. With the charge fixed, the field is set by the charge per unit area on the plates, and moving them apart does not change that.
• Read which case the question describes before writing anything. The two give opposite answers to "what happens to the field".

Worked example: deflecting a beam

Marks that slip away

You've got it