Exam Style Answers 21 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 21 8

Exam-style questions 1

C

2

B

3

a

The foil is positively charged and experiences a force in the same direction as the electric field.   [1]

b

The foil will become bec ome negatively charged

(Direction of field: away away from the positive charge/ towards negative. Force: the charge on an electron is negative therefore force away from the negative plate)  [1] (A: charge on an electron, C: calculated E  as  as V × d , D: calculated d/V calculated d/V only)  [1]

  4

5

6

Five good lines, not touching, good shape [2] [1]

 

and will experience a force in the opposite direction to the field.  [1]

(One mark for three good lines, not touching, good shape)

 

Arrowss in correct direction  Arrow

field strength E = F/Q 

[1]

 = 4.4 × 10−13 N/8.8 × 10−17 C = 5000 N C−1 

[1]

p.d. V = E × d  

[1]

 = 4000 × 0.04 = 160 V 

[1]

a

[1]

separation d = V/E   = 2400/3.0 × 104 = 0.08

  b

m = 8.0 cm 

b

 C

[1]

The sphere would still be attracted to the plate.  [1] The negative charge on the sphere now induces positive charges on the plate.  [1]

 ii ii

The field direction would reverse. 

[1] 

−1

= 1.2 × 10

 V m   [1] The field is directly proportional to the p.d., so doubling the p.d. doubles the field strength. [1] The  T he field strength is inversely proportional to the plate separation, so reducing the separation by a factor of 3 trebles the field strength.  [1] Therefore,  T herefore, the electric field strength is increased by a factor of 6. 

[1]

(But any [0]) indication that the shape changes 9

a

A series of parallel lines between the plates 

[1]

Arrows vertica vertically lly downwards 

[1]

b

Vertically downwards 

[1]

c

6.4 × 10−14 N 

[1]

 d

E = F/q  F/q  = 6.4 × 10−14/1.6 × 10−19 

[1]

 

 

= 400

e

 

1

i

 

[1]

[1]

The positive charge on the sphere induces negative charges on the plate.   [1] The opposite charges attract. 

 

field strength E  = V /d  [1]  [1] = 2400/0.02  [1] 5

  7

a

000 V  

[1]

E = V /d  leading  leading to V  = Ed  = 400 000 × 2.5 −2 × 10   [1] = 10

000 V  

[1]

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

   

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

10 a

Electric field strength is force per unit charge on a stationary charge  [1]

 

per unit positive charge 

[1]

E = V/d  = 5.0 × 106/8.0 × 10−2 

[1]

b

F = EQ  EQ  = (5.0 × 106/8.0 × 10−2) × 1.6 × 10−19   [1]

 

= 1.0 × 10−11 N 

  c

2

b

 

−11

DW = F Dx = 1.0 × 10

 × 8.0 × 10−2 

[1] [1]

d

8.0 × 10−13 J 

[1]

 e

E k = ½  ½mv mv2 = 8.0 × 10−13 

[1]

 

v2 = 2 × 8.0 × 10−13/1.7 × 10−27 

[1]

 

v = 3.1 × 107 m s−1 

[1]

i 

Arrows from the inner electrode to Arrows the outer electrode  [1]

 ii  ii

Lines are closer together together.. 

DV = E Dx = 5.0 × 106 × 1.25 × 10−3  = 6250

c

[1]

work done = 8.0 × 10−13 J 

 

11 a 

V 

[1] [1] [1]

Given that E  = 5.0 × 106 N C−1 (or 5.0 ×  106 V m−1), in a distance of 4.0 µm the potential drops 5.0 × 106 × 4.0 × 10−6 V  [1]

 

potential drop = 20 V 

 

(Other routes are possible possible.) .)

[1]

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