Physics (AS & A Level) · Oct / Nov 2016 · Paper 23 · Question 2
9702
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2(a)[1]
force per unit positive charge [1]
2(b)(i)[1]
time = 5.9 × 10–2 / 3.7 × 107 = 1.6 × 10–9 s (1.59 × 10–9 s) [1]
2(b)(ii)[2]
E = V / d = 2500 / 4.0 × 10–2 = 6.3 × 104 N C–1 (6.25 × 104 or 62500 N C–1) [2]
2(b)(iii)[2]
a = Eq / m or F = ma and F = Eq = (6.3 × 104 × 1.60 × 10–19) / 9.11 × 10–31 = 1.1 × 1016 m s–2 [2]
2(b)(iv)[3]
s = ut + ½at 2 = ½ × 1.1 × 1016 × (1.6 × 10–9) 2 = 1.4 × 10–2 (m) distance from plate = 2.0 – 1.4 = 0.6 cm (allow 1 or more s.f.) [3]
2(b)(v)[1]
electric force ≫ gravitational force (on electron)/weight or acceleration due to electric field ≫ acceleration due to gravitational field [1]
2(b)(vi)[2]
vX–t graph: horizontal line at a non-zero value of vX vY–t graph: straight line through the origin with positive gradient [2]
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