The short answer
Read a field or equipotential map to say which way a charge accelerates and what energy it gains, and scale field, potential and stored charge correctly when a distance or a geometry changes.
Written and checked by GAMSAT tutors — not AI-generated.
Try the reasoning style
We treat forgetting as a failure — a lapse to be patched with reminders and records. Yet a mind that kept everything could not think; it would drown in the undifferentiated noise of every moment it had ever lived. To forget is not so much to lose information as to decide, mostly without our noticing, what was never worth keeping.
The author's argument relies most directly on which unstated assumption?
Pick an option to see how the tutor reasons to the answer — not just whether you were right.
Not quite — the answer is B.
Work backwards from the conclusion: a mind that ‘kept everything’ supposedly ‘could not think.’ That only follows if thinking means leaving most of experience out — so B is the premise the argument quietly rests on. A raises reliability, which the passage never weighs; C contradicts ‘mostly without our noticing’; D smuggles in a claim about intellect the passage never makes. The question rewards finding the hidden premise, not recalling a fact.
Section III hands you unfamiliar apparatus and a column of voltages: which way does a charge move, what energy does it gain? Field is force per charge (F = qE), potential energy per charge (U = qV). Work in potential — W = qΔV, endpoints only, never the path.
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The move: geometry, then the power
Decide the geometry
Point charge, or parallel plates? Between plates the field is uniform: E = V/d everywhere, no inverse square.
Take the power, scale by the ratio
For a point charge, force and field carry 1/r², potential and energy carry 1/r — one power apart. Distance triples: divide E by 9, V by 3, never recompute.
Count the charges
Doubling one charge doubles the force; doubling both quadruples it.
| Distance from the centre (cm) | Potential of the surface (V) | Gap to the next surface (cm) |
|---|---|---|
| 2.0 | 90 | 0.4 |
| 2.4 | 75 | 0.6 |
| 3.0 | 60 | 1.0 |
| 4.0 | 45 | 2.0 |
| 6.0 | 30 | 6.0 |
| 12.0 | 15 | — |
The trap: zero field is not zero potential
Midway between two equal positive charges the fields cancel (E = 0) but the potentials add (V large); between +q and −q the reverse. And a charge accelerates to lower qV, not V, so an electron heads for higher potential.
Worked example
A proton (+e, mass m) and an α-particle (+2e, mass ≈ 4m) cross the same 200 V from rest. Which gains more energy? Which arrives faster?
Capacitance: what is held fixed?
C = Q/V is hardware; U = ½CV², so doubling V quadruples the energy. Then ask what is pinned: triple C with the battery connected and V holds, so Q and the energy triple; disconnected, Q is stranded and both fall to a third.
Check yourself
A small charged bead is fixed inside a vacuum chamber, far from anything else. A probe held 4.0 cm from the bead reads a potential of 90 V and a field strength of 2250 V/m. The probe is moved out to 12.0 cm along the same radial line. It now reads:
Key takeaways
- Force and field go as 1/r², potential and energy as 1/r; for a point charge, E = V/r.
- E can be zero where V is not, and vice versa.
- W = qΔV, endpoints only; charges lower qV, so electrons climb to higher potential.
- U = ½CV²: does the supply pin V, or isolation pin Q?
Practise this with real GAMSAT-style questions
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