Exam Style Answers 23 Asal Physics CB

 

 

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different.

 Coursebook answers Chapter 23 Exam-style questions 1

2

A (B: incorrect calculation of charge (use of V 2), C & D: incorrect calculation of capacitance (use of V  not  not V 2)) 

3

[1]

Three in series, 33 µF 

[1]

Two in parallel, the third in series with the pair 67 µF  [1]

[1]

B (Both 1 and 4 increase the time constant (= CR CR); ); large time constant leads to long decay time)  [1] 6

11 Three in parallel, 300 μF  

Two in series, the third in parallel over the pair, 150 µF  12 a

1 1 1 1  =  10    + 200  + 600  = 100 C ttotal 600 otal

[1] [1]

3

Q = CV = 470 × 10  × 9.0 = 4.2 × 10  C 

[1]

 

4

.03 033 3  × 10 6 = 15 V  V = Q  =  02200 C 

so, combined capacitance = 600  = 60 µF [1] 10

[1]

 

charge stored = 1.5 × 60 × 10 6 = 90 µC  [1]

5 6

2.0  = 4.0 × 10 4 = 400 µF  C = V  = 5000 [1] 1  2 6 2 W  = 2 CV  = 0.5 × 470 × 10  × 12  = 0.034 J  [1]

 

Q1 = +90 µC, Q2 = −90 µC, Q3 = +90 µC, Q4 = −90 µC, Q5 = +90 µC, Q6 = −90 µC

7

W  = 12 QV  = 0.5 × 1.5 × 10 3 × 50 = 0.0375 J  [1]

8

a

−

−

−

Q

−

−

−

b

 

 ½CV  W = ½ CV 2 = 0.5 × 5000 × 10 6 × 242 =  1.44 J  [1]

 

[1]

energy dissipated in lamp =  1.44 J − 0.36 J = 1.08 J 

[1]

[1]

p.d. = 0.90 V across the 100 µF capacitor, capacitor, 0.45 V across the 200 µF capacitor  [1]

 

0.15 V across the 600 µF capacitor 

[1]

13 a 

Time delay, antisu antisurge, rge, antispa antispark, rk, etc.  

[1]

i  R = V   = 9.0/15 × 10−3  I 

[1]

 b

R = 600 Ω 

[1]

 

b

Q = CV = 4700 × 10 6 × 12 = 0.056 C 

[1]

 

c

average current = Q/t Q/t  = 0.056/2.5 =  0.023 A 

[1]

drive current through the resistor   iii  Evidence of using the area  iii

average p.d. = 6.0 V 

[1]

 

R = V/I  = 6.0/0.023 = 260 Ω 

[1]

 iivv  C  = Q/V  = 45 × 10 3/9.0 

  e

−

−

Current is dependent on p.d., which decreases at a non-uniform rate 

 ii  p.d. decreases across capacitor  ii

14 a

[1]

as charge flows off, so less p.d. to

45 ± 5 mC 

=

[1] [1] [1]

−

  [1]

[2]

 

When charge is halved, p.d. is halved  [1] so energy stored =  0.5 × 5000 × 10 6 × 122 = 0.36 J 

Q

Using V = C  

W  = ½  ½CV  CV 2 = 0.5 × 4700 × 10 6 × 122 =  0.34 J  [1]

a

d

10

b

−

−

9

−

 5.0 × 10 3 µF 

[1] [1]

−

i  Q = V  × 4πε0 r = 5.4 × 103 × 4π × 8.85 12 × 10  × 0.20 −

 

=

 1.2 × 10 7 C −

ii  C = Q/V = 2.4 × 10 8 / 5.4 × 103 =  2.2 × 10 11 F = 2.2 pF −

−

1

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020

   

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

b

 or 12 CV 2 = 12  × 1.2 × 10 7 ×  E  = 12 QV  or 5.4 × 103 (or equivalent) 

[1] 

 Q  × frequency = 804 × 10 6 × 50  iiii  I = Qt  = Q   [1]

initial energy = 3.24 × 10 4 J 

[1] 

 

p.d.. after discharge = 15 kV  p.d

[1]

 where V  is  is average p.d.   iiii ii  P = VI  where

−

−

−

 ttherefore, herefore, energy remaining = ½  ½CV  CV 2  11 = ½ × 2.2 × 10  × 15 0002 = 0.0025 J  [1]

=

 eenergy nergy released  0.0324  0.0025 ≈ 0.03 J  c

  15 a

 

−

 

P = 0.24 W 

 b

−

 

=

 4500 V 

 ½QV   ii E  = ½ ii QV  = ½ × 5 × 10

8

−

×

 4500 

Charge stored is halved. 

[1]

 

Current is halved but (av (average) erage) V  is  is unchanged. 

[1]

 

Power is halved.  

[1]

 

(Maximum  [2] if qualitativ qualitative, e, i.e., capacitance reduced, etc.)

17 a

The time constant is the time taken for the charge on a capacitor to fall to 1/e 1/e of its initial value  [1]

 

=

[1] [1]

[1]

 

[1]

−

[1]

[1]

[1] [1]

 12  × 12 × 0.04 

=

[1]

Capacitance is halved. 

[1]

Not sufficient p.d. between sphere and plate  [1] to continue ionising the air  Q V  =  so Q = 4  4π πε0 × V × r  4πε 0 r Q C  = = 4πε 0 r   V  5 ×10 8 Q V  = = i   4 πε 0 r 4π × 8.85 ×10 10 12 × 0.1

c

[1]

 

−

=

 0.040 A 

[1]

 RC  

b

[1]

i

1 1  +  1  =  5  = 500   C ttotal 2000 2000 otal

 

so, combined capacitance =  2000  = 400 µF  5

[1]

−

 c

  = 1.1 × 10 4 J  [1] i  Q is shared, therefore, each sphere has a charge of 2.5 × 10 8 C  [1] −

 

and the p.d p.d.. is halved = 2250 V 

 

thus, ½QV  = ½ × 2.5 × 10 8 × 2250 =  2.8 × 10 5 J  [1]

2

Q = 20 000 µC = 0.020 C 

[1]

 

 iii  Q Q0 exp iii =

Capacitance is the charge stored 

 

per unit potential difference across the capacitor plates.  [1] i  Q = CV  = 67 × 10 6 × 12 

[1]

Q = 804 ≈ 800 µC 

[1]

(CRt ) which leads to −

5

=

[1]

−

 

    − t  = exp   and ln  0.05 ( = − 3.0 ) 100  CR 

 ii  Energy is lost to heating the spheres, ii spheres, as charge moves from one to the other.   [1]

 

[1]

−

−

b

 ii  Q = CV  = 400 × 10 6 × 50  ii −

[1]

16 a

[1]

 

 

t

( 400 ×10 × 250 × 10 )

t = 3.0 s 

−6

3

 =

t 1.0 0  

[1] [1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020