Page 14 - PRE-U STPM PHYSICS TERM 2

Physics Term 2 STPM Chapter 13 Capacitors

ε ε A –t t
r 0
RC
13. (a) C = —––– V = V e , ln V = ln V – –––
d 0 0 RC
1
0.8
(b) C increases Gradient = – ––– = – ––– = – 0.02
RC 40
–8
14. (a) (i) 1.20 × 10 C Q = CV 1
1 C = –––––––––––––– F = 25.0 mF
3
(ii) 6.00 × 10 J U = ––QV (2.0 × 10 )(0.02)
–7
2
1
1
(b) (i) 70 pF C = 10 pF (air) 4. C: U = ––CV , U = ––(4.0C)V = 4U 0
2
2
0
1
1
13 C = 60 pF (dielectric) 2 –t 2 t 13
2
RC
C = C + C 5. D: V = V e , ln V = ln V – –––
0
0
1 2 RC
1 0.4
1 Q 2 Gradient = – ––– = ––––, time constant = RC = 50 s
(ii) 1.03 × 10 J U = –– — RC 20
–6
2 C 1
6. A: Effective capacitance: C = ––C, C = 0.4 C,
Work done on system against electrostatic A 4 B
forces. C = C, C = C
C
D
Q
15. (a) (i) (8 + 6) V = (8)(6.0) + (6)(12.0) 7. C: V= V + V = IR + ––
C
C

R
V = 8.6 V ε ε A
r 0
8. D: C = —–––
(ii) Energy loss d
9. B: Charge is conserved.
1 2 1 2 1 2
= [––C V + ––C V ] – ––(C + C )V (4.0 × 12.0 + 6.0 × 6.0) = (4.0 + 6.0)V
2 1 1 2 2 2 2 1 2

= 0.058 J V = 8.4 V
10. C: With dielectric, capacitance is increased.
(b) (i) (8 + 6)V = (6)(12.0) – (8)(6.0)
Q 2
V = 1.7 V 11. B: C = ε C , U = –––
1 r 0 2C
Q

(ii) Energy loss = 0.56 J 12. B: Charge is conserved, new V = –––—— , some
C + C
1 2
6 energy dissipated as heat.
13. A 14. A 15. B
1. D 2. B 3. C 4. C 5. D 16. C: Induced charges on opposite faces of dielectric
6. B 7. C 8. C 9. D 10. A sets up a reverse field that lowers the electric field
11. (a) Before ball strikes the bat: V constant = 6.0 V in the dielectric.
When ball strikes the bat: V decreases 17. D 18. A 19. C 20. B 21. C
exponentially. 22. C
1
1
1
When ball leaves the bat: V constant. 23. (a) ––– = ––– + ––– , C= 1.71 µF
(b) (i) 0.292 V = 6.0 V, V = 1.75 V C 3.0 4.0
0 e
– —– t (b) Same charge on both capacitors,
–3
(ii) 2.96 × 10 s, V = V e CR
0
12. (a) 6.0 V Q = CV = (1.71 µF)(12.0 V)

(b) V decreases exponentially: Capacitor = 20.5 µC

discharges through R. (c) With dielectric between the plates,
(c) C = 1 µF, R = 5 kΩ capacitance increases.
–3
τ = CR = 5.0 × 10 s is in the same order as time Equivalent capacitance increases.
taken for bullet to travel 10 cm. Charge stored in both capacitors increases.
( —— = 2.5 × 10 s ) 24. (a) (i) 50 µF and 100 µF capacitors are in series.
0.10
–3

Charge of 50 µF

40
(d) V = voltmeter reading after Y breaks = charge on 100 µF capacitor
-t = 700 µC
= (6.0)e . Calculate t.
CR
(ii) Equivalent capacitance of 200 µF capacitor
(e) Discharge of capacitor through voltmeter
and 150 µF capacitor (in parallel) is 350 µF
negligible.
Charge on 350 µF capacitor = (350 µF)V
STPM PRACTICE 13 = 700 µC
Hence V = 2.0 V
1. A: Charge is conserved,
Charge on 200 µF capacitor
Q = constant
= (200 µF)(2.0 V) = 400 µC
2. D: As V across the capacitor increases, V, across R 1 Q 1 1 1
decreases to zero. (b) — = V = — = (700 µC)(—– + —–– + —––)
C XY C 50 350 100
1
1
1
3. C: –– = ––– + –––, R = 2.0 kΩ = 23.0 V
R 3.0 6.0
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13 Physic T2.indd 79 10/18/18 3:18 PM