Physics (AS & A Level) · Oct / Nov 2016 · Paper 41 · Question 5
9702
As printed in the paper
The question as text
Two small solid metal spheres A and B have equal radii and are in a vacuum. Their centres are 15 cm apart. Sphere A has charge +3.0 pC and sphere B has charge +12 pC. The arrangement is illustrated in Fig. 5.1. 15 cm 5.0 cm P sphere A charge + 3.0 pC sphere B charge + 12 pC Fig. 5.1 Point P lies on the line joining the centres of the spheres and is a distance of 5.0 cm from the centre of sphere A.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[2]
in an electric field, charges (in a conductor) would move no movement of charge so zero field strength or charge moves until F = 0 / E = 0 [2]
or charges in metal do not move (B1) no (resultant) force on charges so no (electric) field (B1)
5(b)[3]
at P, EA = (3.0 × 10–12) / [4πε0(5.0 × 10–2) 2] (= 10.79 N C–1) at P, EB = (12 × 10–12) / [4πε0(10 × 10–2) 2] (= 10.79 N C–1) or (3.0 × 10–12) / [4πε0(5.0 × 10–2) 2] – (12 × 10–12) / [4πε0(10 × 10–2) 2] = 0 or (3.0 × 10–12) / [4πε0(5.0 × 10–2) 2] = (12 × 10–12) / [4πε0(10 × 10–2) 2] (M2) fields due to charged spheres are (equal and) opposite in direction, so E = 0 [3]
5(c)[2]
potential = 8.99 × 109 {(3.0 × 10–12) / (5.0 × 10–2) + (12 × 10–12) / (10 × 10–2)} = 1.62 V [2]
5(d)[3]
½mv2 = qV EK = ½ × 107 × 1.66 × 10–27 × v 2 qV = 47 × 1.60 × 10–19 × 1.62 v 2 = 1.37 × 108 v = 1.2 × 104 m s–1 [3]
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.