Model qualitatively and quantitatively the electric field, including direction and shape, produced between parallel charged plates and the potential difference, using E = V/d

Field strength:

$E = \frac{V}{d} = \frac{60}{0.015} = 4.0 \times 10^3$ V/m.

Direction: the electric field points from the positive plate to the negative plate, so vertically downward (upper plate positive, lower plate negative).

The field is treated as uniform because in the central region between two large, closely spaced parallel plates the field lines are straight, parallel, and evenly spaced, indicating constant magnitude and direction. This is an idealisation; near the edges the lines curve outward (fringing), but for a test charge placed well away from the edges the uniform approximation is excellent.

Markers reward the SI conversion, the direction with reasoning (positive to negative), and the explicit statement of the uniform-field assumption.

The electric field strength between parallel plates is $E = V/d$. Holding $V$ constant and halving $d$ doubles the field strength: $E_{new} = V / (d/2) = 2V/d = 2 E_{old}$.

A doubled field means twice the force on any test charge in the gap, and twice the acceleration of a charged particle placed at rest in the field.

Markers reward the algebraic argument and a statement of the physical consequence (force or acceleration doubles).

Convert to SI units first: $d = 2.0\ \text{cm} = 2.0 \times 10^{-2}\ \text{m}$.

$E = \dfrac{V}{d} = \dfrac{120}{2.0 \times 10^{-2}} = 6.0 \times 10^{3}\ \text{V/m}$.

Marks: one for converting $\text{cm}$ to $\text{m}$ before substituting, one for the correct answer $E = 6.0 \times 10^{3}\ \text{V/m}$ with the unit.

The field points from the positive plate to the negative plate, so it points from left to right (left to right, away from $+$, toward $-$).

If the polarity is reversed, the right plate becomes positive and the left plate becomes negative, so the field direction reverses to point from right to left. The field lines remain straight and evenly spaced in the central region; only the direction (arrowheads) flips.

Marks: one for the correct initial direction (positive to negative), one for correctly stating the field reverses on reversing the polarity.

In the central region, well away from the edges of the plates, the field lines are straight, parallel and evenly spaced, pointing from the positive plate to the negative plate: this is the uniform field where $E = V/d$ applies at every point.

Near the edges of the plates the field lines curve outward (bow away from the straight-line path); this is the fringing or edge effect, where the field is weaker and non-uniform.

Outside the plates (beyond the edges, or above/below an infinite-plate idealisation) the field is essentially zero.

Marks: one for describing the uniform central region, one for describing fringing at the edges, one for noting the field is (approximately) zero outside the plates.

(a) Field strength from the force. $E = \dfrac{F}{q} = \dfrac{4.5 \times 10^{-5}}{3.0 \times 10^{-9}} = 1.5 \times 10^{4}\ \text{V/m}$.

(b) Potential difference. $V = Ed = (1.5 \times 10^{4})(8.0 \times 10^{-3}) = 1.2 \times 10^{2}\ \text{V}$.

Marks: one for $E = F/q$ with substitution, one for $E = 1.5 \times 10^{4}\ \text{V/m}$, one for $V = Ed = 1.2 \times 10^{2}\ \text{V}$ with correct unit conversion of $d$.

(a) Before: $E = \dfrac{9.0}{1.0 \times 10^{-3}} = 9.0 \times 10^{3}\ \text{V/m}$. After: $E = \dfrac{9.0}{3.0 \times 10^{-3}} = 3.0 \times 10^{3}\ \text{V/m}$, a factor of $3$ smaller, since $V$ is held fixed by the battery while $d$ triples.

(b) If the plates are disconnected, the charge $Q$ on them is fixed. Since $C = \varepsilon_0 A / d$, increasing $d$ decreases the capacitance $C$, and because $Q = CV$ is fixed, $V$ must rise as $C$ falls. Because $E = \sigma/\varepsilon_0$ depends only on the (unchanged) surface charge density, $E$ stays constant even though $V$ increases and $d$ increases, consistent with $E = V/d$.

Marks: one for both field values in part (a) with correct unit conversion, one for identifying that $Q$ (hence $\sigma$) is fixed when disconnected, one for concluding $E$ stays constant while $V$ rises to match the larger $d$.

(a) The graph is a straight line through the origin, so $V$ is directly proportional to $d$ ($V \propto d$), as expected from $E = V/d$ rearranged to $V = Ed$ with $E$ held constant.

(b) Gradient $= \dfrac{\Delta V}{\Delta d} = \dfrac{(400 - 100)}{(8.0 - 2.0) \times 10^{-3}} = \dfrac{300}{6.0 \times 10^{-3}} = 5.0 \times 10^{4}\ \text{V/m}$.

(c) The gradient of a $V$-versus-$d$ graph (at fixed $E$) equals the field strength $E$ itself, since $V = Ed$ is a straight line through the origin with gradient $E$. So $E = 5.0 \times 10^{4}\ \text{V/m}$.

Marks: one for identifying direct proportionality (line through the origin, $V \propto d$), one for a correctly computed gradient with the unit V/m, one for correctly identifying the gradient as the field strength $E$, one for stating $E = 5.0 \times 10^{4}\ \text{V/m}$.

Band-6 plan. (1) State the method: measure $V$ from the supply and $d$ with a ruler/vernier, compute $E = V/d$; note this only works in the central region. (2) Identify at least two genuine sources of error (fringing, non-parallel/misaligned plates, parallax in measuring $d$, supply voltage fluctuation, air breakdown at high $V$) and explain the effect of each on the result. (3) Evaluate which errors are systematic versus random, and suggest at least one improvement. (4) Reach a judgement on the reliability of the method.

Model answer. The student should record the supply voltage $V$ directly from a calibrated meter and measure the plate separation $d$ with a vernier caliper or travelling microscope for precision, then calculate $E = V/d$. Crucially, any test-charge measurement (of the force on the charge, to cross-check $E$ via $F = qE$) must be taken well away from the edges of the plates, in the central region, because the field there is uniform; measurements taken near the edges would sample the weaker, non-uniform fringing field and would not represent the "true" $E = V/d$ value.

Several sources of error affect the investigation. Fringing near the edges is not itself a problem for a properly centred measurement, but if the test charge is not carefully centred it introduces a systematic error, since the weaker fringing field there would always cause $E$ to be underestimated (any accidental edge placement biases $E$ low). Non-parallel or misaligned plates introduce a systematic error because $d$ then varies across the gap, so a single ruler measurement is not the true separation everywhere. Parallax error in reading $d$ from a ruler is a random error that can be reduced by using vernier calipers and taking repeated readings. Fluctuation or drift in the DC supply voltage is a random error reducible by using a regulated supply and taking the reading at the same instant as the force measurement. At very high $V$ for small $d$, partial ionisation of the air (approaching the $\sim 3 \times 10^6\ \text{V/m}$ breakdown field) could locally distort the field and is best avoided by keeping the applied field well below breakdown.

Overall, provided the student measures well within the central region, uses precision instruments for $d$, and a regulated, verified supply for $V$, the method $E = V/d$ gives a reliable and accurate field strength; the main residual risk is a systematic underestimate from any edge or alignment effects, best controlled by centring the measurement and checking plate parallelism.

Marker's note: the top band explicitly separates systematic from random error sources, ties each error to its physical cause (fringing, misalignment, parallax, supply drift), and closes with a reasoned judgement on reliability rather than merely listing possible errors.

Band-6 plan. (1) State the claim is false and give the correct picture: uniform in the centre, weaker (not stronger) only at the edges, not near the plate surfaces along the central axis. (2) Use $E = V/d$ to argue why: $E$ has a single value independent of position within the uniform region. (3) Explain physically why it is uniform (field-line spacing constant = constant magnitude). (4) Address where the field genuinely does change (fringing at the edges, not "close to the plates" in the central region). (5) Conclude with a corrected statement.

Model answer. The claim is incorrect. Between two large parallel plates, the field in the central region - which includes points close to either plate surface, provided they are away from the edges - is uniform: the field lines are straight, parallel and evenly spaced throughout the gap, meaning the field has the same magnitude and direction everywhere in that region, not just in the middle. This is captured by $E = V/d$: for a fixed $V$ and $d$, this equation gives a single value of $E$ that applies at every point between the plates, with no dependence on how far a point is from either plate. If the field really were stronger near the plates, $E$ would have to depend on position, and the simple relationship $E = V/d$ would not hold; instead, a full description would need a position-dependent field.

The place where the field genuinely is non-uniform, and does depart from $E = V/d$, is not "close to the plates" in the sense of vertical distance from a plate, but close to the ends (edges) of the plates, where fringing occurs: there the field lines curve outward and the field is weaker than the central value, not stronger. Outside the plates entirely, the field is essentially zero for large, closely spaced plates.

So the correct statement is the opposite in structure to the claim: the field is constant (uniform) throughout the central region between the plates, including near either plate surface, and only becomes weaker and non-uniform near the edges of the plates - it is never stronger anywhere in the ideal parallel-plate field.

Marker's note: the top band explicitly refutes the "close to the plates" framing (a common confusion with a point-charge or point-source field, where strength genuinely does fall off with distance), correctly locates the real non-uniformity at the edges rather than near the plate surfaces, and uses $E = V/d$ as evidence that the field cannot depend on position within the uniform region. Simply asserting "the field is uniform" without addressing why the claim's edge/distance reasoning is wrong caps in the middle band.