Uniform electric fields
Introduced| English | Meaning |
|---|---|
| uniform field/ˈjuːnɪfɔːm fiːld/ | A field with the same magnitude and direction at every point in the region considered. |
| potential difference/pəˈtenʃl ˈdɪfrəns/ | The energy transferred per unit charge between two points. |
| parabola/pəˈræbələ/ | The curve formed by points equally distant from a fixed point and a fixed straight line. |
| projectile/prəˈdʒektaɪl/ | An object moving after being launched, typically modelled under gravity with air resistance neglected. |
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
Two plates $0.050\ \text{m}$ apart have $200\ \text{V}$ between them. What is the field strength?
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$E = \dfrac{V}{d} = \dfrac{200}{0.050} = 4000\ \dfrac{\text{V}}{\text{m}}$.
A uniform field between two plates points:
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The field runs from high potential (+) toward low potential (−).
The unit V/m is the same as N/C.
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Both describe electric field strength; $\dfrac{\text{V}}{\text{m}} = \dfrac{\text{N}}{\text{C}}$.
Which changes increase the field strength between two charged parallel plates? Select all that apply.
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E = V/d depends only on those two. No current flows once the plates are charged, so a series resistor drops no voltage and changes nothing.
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
Uniform electric field lab
Follow how a charge behaves between parallel plates.
A charge in a uniform field has a constant acceleration.
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The force $qE$ is constant, so $a = \dfrac{qE}{m}$ is constant — like free fall in gravity.
A charge entering a uniform field at right angles follows a:
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Constant sideways speed plus steady acceleration across the field gives a parabolic path, like a projectile.
Worked example: an electron crossing the gap
- Two plates in a vacuum are $0.041\ \text{m}$ apart with $250\ \text{V}$ between them. An electron is released from rest at the negative plate. Find the field, the acceleration, the time to cross, and the kinetic energy on arrival.
- $E = V/d = 250/0.041 = 6.1 \times 10^3\ \text{V/m}$, and $F = eE = 9.8 \times 10^{-16}\ \text{N}$, so $a = F/m_{\text{e}} = 1.07 \times 10^{15}\ \text{m/s}^2$.
- From rest with constant acceleration, $d = \tfrac12 at^2$, so $t = \sqrt{2d/a} = 8.8 \times 10^{-9}\ \text{s}$.
- Kinetic energy: use energy, not kinematics. $E_{\text{k}} = qV = (1.60 \times 10^{-19})(250) = 4.0 \times 10^{-17}\ \text{J}$, giving $v = 9.4 \times 10^6\ \text{m/s}$, which the kinematic route confirms.
Worked example: the balanced oil drop
- An oil drop of mass $2.6 \times 10^{-15}\ \text{kg}$ carrying $-4.8 \times 10^{-19}\ \text{C}$ hangs stationary between horizontal plates $2.0\ \text{cm}$ apart. Find the p.d., and say which plate is positive.
- Two forces act, not one: the electric force and the weight. "Electric force only" is the standard wrong answer. Stationary means they are equal and opposite:
- The drop is negative, so the force on it is opposite to the field. For that force to be upwards, the field must point down, so the top plate is positive.
- The charge here is $3e$: three excess electrons. That whole-number multiple is Millikan's result.
An electron is released from rest and crosses a p.d. of 250 V. What is its kinetic energy on arrival, in units of 10^-17 J? (e = 1.60e-19 C)
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Ek = qV = 1.60e-19 x 250 = 4.0e-17 J, with no kinematics needed. The energy route is almost always quicker than finding a, then t, then v.
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".
A charged oil drop hangs stationary between two horizontal plates. Which forces act on it?
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"Electric force only" is the standard distractor. Stationary means the two forces balance, which is what makes the drop a measuring device for its own charge.
Two parallel plates are moved further apart. Match each case to what happens to the field between them.
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With the charge fixed the field is set by the charge per unit area, which moving the plates does not change. Read which case the question describes before answering.
Worked example: deflecting a beam
- An electron travelling horizontally at $2.0 \times 10^7\ \text{m/s}$ enters the field between horizontal plates $18\ \text{mm}$ apart with $400\ \text{V}$ across them. The plates are $30\ \text{mm}$ long. Find the deflection as it leaves.
- Treat the two directions separately, as with any projectile. Horizontally the motion is unaffected, so the time in the field is $t = 0.030/(2.0 \times 10^7) = 1.5 \times 10^{-9}\ \text{s}$.
- Vertically, $a = eV/(m_{\text{e}}d) = 3.9 \times 10^{15}\ \text{m/s}^2$, so the deflection is $y = \tfrac12 at^2 = 4.4\ \text{mm}$.
- The horizontal speed sets the time, and the field sets the acceleration. Mixing them is the usual error.
Put the steps of finding the deflection of an electron crossing a uniform field in order.
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The horizontal speed sets the time and the field sets the acceleration. Treating it as a projectile is the whole method.
Marks that slip away
- A stationary charged drop has two forces on it, the electric force and its weight. Say both.
- For a negative charge, the force is opposite to the field, which is what decides which plate is positive.
- The field between plates depends only on $V$ and $d$. A series resistor changes nothing, since no current flows.
- For a deflection, split the motion: constant velocity along, constant acceleration across, and the time comes from the horizontal distance.
You've got it
- between parallel plates the field is uniform, $E = V/d$, directed from the higher-potential plate to the lower; only changing $V$ or $d$ changes it
- a charge feels a constant force $qE$ and so a constant acceleration; released from rest it gains $E_{\text{k}} = qV$
- entering at right angles it follows a parabola: constant speed along, constant acceleration across, with the time set by the horizontal distance
- a stationary drop balances the electric force against its weight, $qV/d = mg$, and the sign of the charge decides which plate is positive