Friday, December 04, 2020
Wednesday, December 02, 2020
Electric Fields Revision Questions
3. (a) appropriate shape; lines perpendicular to
and touching plate and sphere; (2)
arrows towards negative sphere (1) 3
(b) (i) By moments, e.g F cos 20 = W sin 20 / by
triangle of forces /
by resolution of forces / other suitable method; i.e. justification needed
(1)
F = 1.0 × 10–5 tan 20; =
1.0 × 10–5 × 0.364; (= 3.64 × 10–6 N) (2)
triangle of forces gives W/F = tan 70, etc (1) 3
(ii) E
= F/Q; = 3.64 × 10–6 / 1.2 ×
10–9 = 3.0 × 103;N
C–1 / V m–1 3
(c) E
= (1/4πεo)Q/r2; 3.0 × 103 = 9 × 109 × 1.2 × 10–9/r2;
(2)
or use F = (1/4πεo)Q2/r2; r2 = 3.6 × 10–3 giving r = 6 × 10–2 (m) (1) 3
(d) field line sketch minimum of 5 lines symmetrical
about line joining
centres with arrows; (1)
Fig 1 sketch matches RHS of Fig 2/plate analogous to mirror/AW
relating to symmetry (1) 2
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(i) Suitable recognisable pattern around (not
just between) the charges B1
Quality mark: symmetry, spacing, lines joined to charges B1
Consistent arrows toward B on some lines B1
(ii) Use
of E = (1/4πε0)Q/r2 C1
Sum of two equal terms
E = 2 × 9 × 109 × 1.6 × 10–19 / (2.0 × 10–10)2 C1
E = 7.2 × 1010 N C–1 or V m–1 A1
(iii) The
separation between the ions because this has an effect on the
breaking force. (Allow the size of ionic ‘charges’) B1
[7]
y12 Electricity revision question on Christmas tree bulbs and 2 lamps
1. (a) either (If in parallel) when one bulb fails, other
bulbs stay on
or (If in parallel) can
identify which bulb has failed; (1) 1
(b) (i) P = VI (1)
0.5 = 240 I
I = 2.1 × 10–3 A 1
s.f. in answer (–1) once only (1) 2
(ii) R
= V/I (1)
= 240/(2.1 × 10–3)
= 1.14 × 105 Ω or 1.15
× 105 Ω ans
accept (1.1 to 1.2) × 105 Ω. (1) 2
(iii) A
= ρ l / R (1)
= 1.1 × 10–6 × 6.0 ×
10–3 / (1.14 × 105)
(= 5.79 × 10–14 m2)
A = πr2 (1)
5.79 × 10–14 = πr2 so r = 1.4 × 10–7 m (1) 3
(iv) filament
too thin / fragile to be manufactured / used without damage;
allow ecf from (iii). (1) 1
4
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Tuesday, December 01, 2020
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