4 Electricity and Magnetism Chapter 1 Electrostatics

4 Electricity and Magnetism Chapter 1 Electrostatics

1 Electrostatics

New Senior Secondary Physics at Work 8 Ó Oxford University Press 2010

4 Electricity and Magnetism Chapter 1 Electrostatics

Practice 1.1 (p. 18)

1 D

2 D

3 D

4 (a) False

(b) True

(d) True

5 (a) By Coulomb’s Law,

F =

= 3.60 ´ 10–3 N

(b) By Coulomb’s Law,

= 20.2 m

6 When rubbed, the plastic comb is charged by friction. Suppose the comb carries a positive charge. When the comb is put close to hairs, a negative charge is induced on hairs, on the side close to the comb; a positive charge is induced on hairs, on the side away from the comb. Since the negative induced charge on hairs is closer to the comb, the attraction between hairs and the comb is larger than the repulsion. Therefore, the comb attracts hairs. The case is similar when the comb is negatively charged.

7 This is because as soon as the charges appear on the metal rod after rubbing, they are conducted away to the earth via the human body.

A metal rod can be charged by rubbing when it is insulated from the human body and the surroundings.

8 (a) The two spheres repel each other.

(b) Sphere B is earthed due to the metal thread. A negative charge is induced on sphere B on the side facing sphere A and a positive charge is induced on the side away from sphere A. Since sphere A is closer to the negative induced charge than the positive one, sphere B is attracted towards sphere A.

If the attractive force is strong enough, the two spheres will touch each other. Both spheres will be earthed and become neutral. Then they will swing back to their original positions.

(a) Consider the vertical direction.

T cos 10° = mg = 0.0210 = 0.2 N

T == 0.203 N

(b) Fe = T sin 10° = 0.203 sin 10° = 0.0353 N

QB =

= -7.85 ´ 10-7 C

Practice 1.2 (p. 27)

1 D

2 C

3 (a) Electric field strength

= 2.30 ´ 104 N C-1 (towards left)

(b) Electric field strength

= –3.60 ´ 104 N C-1 (towards left)

= 3.60 ´ 104 N C-1 (towards right)

(c)

Electric field strength E1

= 3.03 ´ 103 N C-1

tan q ==

q = 8.43°

4 (a) E ==

= 1.81 ´ 106 N C-1

(b) The field strength will increase to four times of its original value.

5 (a)

(b)

(c)

(d)

6 There will be an electric force acting on the small positive charge along the direction of electric field line. If the charge is replaced with a small negative charge, the electric force acting on it will be opposite to the direction of electric field line.

Practice 1.3 (p. 37)

1 C

2 C

Since V = -Ex,

potential difference

= -EDx

= -1.8 ´ 105 ´ (0.04 - 0.02)

= -3600 V

The potential difference between the points at

x = 2 cm and x = 4 cm is 3600 V.

3 (a) By V =,

r === 0.270 m

(b) Potential energy = q ´ V

= -2 ´ 10-8 ´ 1 ´ 103

= -2 ´ 10–5 J

4 (a) Electric potential at X

= 0

(b) Distance between Q1 and X

== 0.9487 m

Electric potential at X

= -710 V

5 Work done = potential energy difference

= QB ´ (VX - VB)

= QB ´

= 6.74 ´ 10–5 J

6 (a) Potential difference between A and B

= VA - VB

= 8.99 ´ 103 V

(b) Electric field strength between two plates

=== 1.356 ´ 104 N C-1

Since V = -Ex,

potential difference between A and B

= -ErAB

= -1.356 ´ 104 ´ 0.03

= -407 V

The potential difference between A and B is 407 V.

= VA - VB

= –2.40 ´ 104 V

The potential difference between A and B is 2.40 ´ 104 V.

Revision exercise 1

Multiple-choice (p. 41)

1 B

2 B

3 C

4 A

5 C

6 C

7 D

8 B

9 C

10 D

As the potential at X is 0, P and Q must have charge in different signs.

Let d be the distance PX.

Let q be the charge of Q.

= 0

d = 0.45 m

= 45 cm

11 A

Electrostatic force acting on charge Q

= QE ==

12 D

13 (HKCEE 2003 Paper II Q44)

14 (HKALE 2003 Paper II Q23)

15 (HKCEE 2004 Paper II Q27)

16 (HKALE 2004 Paper II Q16)

17 C

When the charge is moved from P to Q, it is moved along the equipotential line which is perpendicular to the electric field lines.

Therefore, the change in PE from P to Q is zero.

Change in PE when the charge is moved from Q to R

= PER - PEQ

= qEd

= 4 ´ 10-6 ´ 10 ´ 3

= 120 ´ 10-6 J

= 120 mJ

Total change in PE from P to R

= 0 + 120 = 120 mJ

Conventional (p. 43)

1 (a) Sphere A carries a negative charge. (1A)

Since like charges repel, the repulsion between spheres A and B implies that sphere A carries the same type of charge as sphere B, i.e. a negative charge. (1A)

(b)

(Correct weight.) (1A)

(Correct tension.) (1A)

(Correct electrostatic force: repulsion.)

(1A)

(The forces should balance each other.)

The student is incorrect. (1A)

2 (a) A is negatively charged. (1A)

C is positively charged. (1A)

(b)

(Correct free-body diagram for A.) (1A)

(Correct free-body diagram for B.) (1A)

(Correct free-body diagram for C.) (1A)

(ii) If A and B are interchanged, B will deflect to the right and (1A)

C will deflect to the left. (1A)

The deflection of A depends on its distance from B and C. (1A)

3 (a) Charge is accumulated. (1A)

When the car runs on a road, its tyres and body are charged by friction with the road and the air respectively. (1A)

(b) When the driver steps on the ground with his hand touching a car door, the car is earthed through the driver’s body. (1A)

Electrons flow through the driver from/to the car to/from the earth (depending on the type of charge of the car). (1A)

The flow of electrons through the driver causes an electric shock. (1A)

(c) This can be prevented by earthing, e.g. connecting the car and the earth with a metal chain at the back of the car. (1A)

4 (a) & (c)

For (a):

(Correct axes and labels.) (1A)

(Straight line passing through the origin.) (1A)

(Correct slope.) (1A)

For (c):

(Straight line passing through the origin.) (1A)

(New slope = half the slope of line in (a).) (1A)

(b) Potential difference between the two plates

= -Ed

= -(-2.8 ´ 105) ´ 0.06

= 1.68 ´ 104 V (1A)

5 (a) Sum of energy

= KE + electric PE

= (1M)

= 5.18 ´ 10–4 J (1A)

(b) At the position P stops moving towards Q, the KE of P is zero. By conservation of energy, the electric PE of P is equal to

5.18 ´ 10–4 J. (1M)

Let d be the distance of P from Q at that moment.

= 5.18 ´ 10–4 (1M)

d =

= 0.01042 m

» 1.04 cm (1A)

= (1M)

= 5.18 ´ 10–4 ´

= 4.97 ´ 10–2 N (1A)

6 (a) When lightning hit the lightning rod, the metal ball would hit bell A first. (1A)

When lightning hit the lightning rod, bell A would be negatively charged. Positive induced charges would appear on the surface of the metal ball facing bell A, while negative induced charges would appear on the other side. (1A)

Since unlike charges attract and like charges repel (1A)

and induced positive charges would be closer to the bell A, the metal ball would be attracted towards bell A and hit bell A first. (1A)

(b) (i) Sparks are produced as charges jump from bell A to the ball. (1A)

(ii)

(Correct shape of electric field lines.) (1A)

(Correct direction of electric field lines.) (1A)

7 (a) (i) Electron is transferred from one material to the other. (1A)

(ii) When the charge built up is large enough, sparks can be formed. (1A)

If sparks ignite petrol vapour, explosion results. (1A)

Therefore, it is very dangerous.

(iii) A metal chain can be used to connect the pipe and the earth to prevent the build-up of charge. (1A)

(b) The granules gain the same type of charge when they pass the funnel. (1A)

Since like charges repel, the granules repel each other. Therefore, some of them are pushed out sideways and miss the container. (1A)

8 (HKALE 2002 Paper I Q3)

9 (a) (i) It must have units so that dimensions on each side of any equation involving e 0 balance. (1A)

(ii)

(Curve of correct shape.) (1A)

(Not touching either axis.) (1A)

(b) (i)

(Correct direction at X.) (1A)

(Correct direction at Y.) (1A)

(ii) Electric field strength at X

= (1M)

= -1.92 ´ 108 N C-1

The magnitude of the electric field strength at X is 1.92 ´ 108 N C-1. (1A)

(Correct polarity.) (1A)

(ii)

(Finite start point.) (1A)

(Upward curve, correct shape.) (1A)

10 (a) (i) & (ii)

For (i):

(Correct curve shape.) (1A)

(Asymptotic both axes.) (1A)

For (ii):

(Correct similar curve shape, always above graph 1.) (1A)

(b) (i)

(Three arrows are all horizontal and all from right to left.) (3 ´ 1A)

(ii) At Y. (1A)

(Straight line through the origin.) (1A)

(ii)

(Correct polarity.) (1A)

(iii) Potential difference between the plates

= Ed (1M)

= 7.0 ´ 103 ´ 0.012 (1M)

= 84 V (1A)

Physics in articles (p. 47)

1 (a) (i) When a positively-charged rod is put near the electroscope, a negative charge is induced on the metal plate. (1A)

A positive charge is induced in the metal support and the gold leaf so that the positive charge is further from the positively-charged rod. (1A)

Because of the repulsion between like charges, the gold leaf will rise. (1A)

(ii) The gold leaf will drop. (1A)

(b) (i) The gold leaf will rise. (1A)

(ii) The gold leaf will drop. (1A)

2 (a) The droplets of spray with the same charge repel each other. Therefore tiny droplets will not accumulate as a large droplet of spray. (1A)

The skin is originally neutral. Charge is induced on the skin, which is opposite to the charge of the droplets. (1A)

Droplets are attracted to the skin and stick on it because of the induced charge. (1A)

(b) Only a little amount (about 20 millilitres) of new spray is wasted in each spray. (1A)

The droplets will not stick to the skin. (1A)

New Senior Secondary Physics at Work 8 Ó Oxford University Press 2010