Forces, components and equilibrium
Resolve forces, balance moments and distinguish force from torque.
Gravitational and electric field questions
Distinguish field strength from potential; use centre distances, charge signs and circular orbits.
Practise
Changing distanceOutside a spherical planet, the distance from its centre increases from r to 2r. What is the new gravitational field strength as a fraction of its original value?
Hint
Field strength is proportional to 1/r².
Worked solution
- Rule: 1/4.
- gnew/gold = r²/(2r)² = 1/4.
Question type: Mixed practice
Report a problem
Measure radius from the centre
Outside a spherical planet, g = GM/r². For height h above a planet of radius R, use r = R + h. Gravitational field strength points towards the planet; its magnitude is positive. In a circular orbit, gravity provides centripetal force, giving speed √(GM/r).
Field strength and potential are different quantities
Gravitational potential is energy per unit mass: V = −GM/r, taking zero at infinity. It is measured in J kg⁻¹. Gravitational field strength is force per unit mass, measured in N kg⁻¹. Do not use the inverse-square field equation to calculate potential.
Keep the charge sign
For a point charge in vacuum, field magnitude E = k|Q|/r² and signed potential V = kQ/r. Field points away from positive charge and towards negative charge. Between parallel plates, E = ΔV/d away from the positive plate, ignoring edge effects. Force F = qE reverses direction for a negative charge.
What this practice covers
Selected spherical gravitational fields, circular orbital speeds, point-charge electric fields and potentials, and forces in uniform fields. No field superposition, magnetic fields, escape speed, equipotential plotting or charged-particle trajectories are included.
A-level relevance
AQA 7408: 3.7.2.1–4; 3.7.3.2–3 (selected); OCR A H556: 5.4.1–4; 6.2.1–4 (selected). Tasks cover the selected skills listed above.