Physics

1. 1.

   Outside a spherical planet the gravitational field at radius R is 8.0 N kg^-1. Find the field at 2R and the ratio of potential there to potential at R, taking potential zero at infinity.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Doubling distance divides the field by four but divides the negative potential magnitude by two.
   2. The potential becomes less negative, approaching zero from below.

   Marking points

   • Field magnitude varies as 1/r^2.
   • Field at 2R is 2.0 N kg^-1 inward.
   • Potential varies as -1/r, so V(2R)/V(R) = 1/2.

   Examiner tip: Do not apply the inverse-square rule to potential.

2. 2.

   A charge +3.00 microcoulomb moves slowly from potential 20.0 V to 80.0 V. Find its change in electric potential energy and work done by the electric field.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. A positive charge gains potential energy when moved to a higher potential.
   2. The electric field's work equals the negative of its potential energy change; the external agent supplies positive work in slow motion.

   Marking points

   • delta U = q(V_final - V_initial).
   • delta U = +0.000180 J.
   • Field work is -0.000180 J (3 significant figures).

   Examiner tip: State whose work is requested: field work and external work have opposite signs.

3. 3.

   A satellite orbits a spherical planet at radius 7.00e6 m from its centre. GM = 3.99e14 m^3 s^-2. Find circular orbital speed and period, ignoring other bodies, to 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The orbit radius is measured from the centre, not from the surface.
   2. Divide circumference by unrounded speed to avoid compounding rounding in the period.

   Marking points

   • GMm/r^2 = mv^2/r, so v = sqrt(GM/r).
   • v = 7550 m s^-1 (3 significant figures).
   • T = 2pi r/v = 5830 s (3 significant figures).

   Examiner tip: Orbital speed is not escape speed; it lacks the extra sqrt(2).

4. 4.

   Two point charges +2.00 microcoulomb and -2.00 microcoulomb are 0.400 m apart in vacuum. Find potential and electric field at their midpoint. Use k = 8.99e9 N m^2 C^-2.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Potential is scalar, so use charge signs in the sum; field is vector, so compare directions.
   2. At the midpoint the positive charge pushes a test positive charge towards the negative charge, which attracts it in the same direction.

   Marking points

   • Scalar potentials cancel, giving V = 0 V.
   • Each field has magnitude kq/(0.200)^2 = 449500 N C^-1.
   • Fields add towards the negative charge: E = 8.99e5 N C^-1.

   Examiner tip: Zero potential does not imply zero electric field.

5. 5.

   A spherical planet has surface radius R = 6.00e6 m and surface g = 10.0 m s^-2. Find minimum surface launch speed to reach radius 3R with zero speed there. Ignore atmosphere and rotation. Compare with escape speed, to 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Use gravitational potential -GM/r because g is not constant over this distance.
   2. At the target kinetic energy is zero; escape replaces the target potential by zero at infinity.

   Marking points

   • GM = gR^2; energy gives v^2/2 = GM(1/R - 1/(3R)).
   • v = sqrt(4gR/3) = 8940 m s^-1 (3 significant figures).
   • Escape speed sqrt(2gR) = 1.10e4 m s^-1; finite-height speed is sqrt(2/3) of escape speed.

   Examiner tip: mgh is a near-surface approximation and is unsuitable for a height of 2R.

6. 6.

   Point charges +Q at x = 0 and +4Q at x = 0.600 m lie in vacuum, Q > 0. Find the finite point on the x-axis where net field is zero and explain why net potential is not zero there with zero at infinity.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Outside the interval the two fields point in the same direction, so squared-equation roots there must be rejected.
   2. The cancellation point lies closer to the weaker charge; potentials add without vector cancellation.

   Marking points

   • Cancellation is possible only between the charges, where their fields oppose.
   • Q/x^2 = 4Q/(0.600 - x)^2 gives 0.600 - x = 2x, so x = 0.200 m.
   • V = kQ/0.200 + 4kQ/0.400 = 15.0kQ m^-1 > 0; both contributions are positive.

   Examiner tip: Check field directions before taking square roots of the magnitude equation.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.