© Pearson Education Asia Limited 2015 Active Physics DSE Traps (Sample)
4. Electricity and Magnetism
Ï Covers all HKDSE Exam Report comments
Ï Includes a Tip and a Trap designed to cater for each weak point mentioned in the Report • Each trap consists of a question and an attempted answer • Students are asked to determine whether the attempted answer scores full marks
Ï Helps students score higher in Exam © Pearson Education Asia Limited 2015 Chapter 20 Electrostatics © Pearson Education Asia Limited 2015 Tip 20.1 Electric force
Remember that the electric force F between two charges is inversely proportional to the square of their separation r , i.e. F ∝ 1 . r 2 (DSE-14-1A Q21) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
The electric force between two point charges is F . If the separation of the two charges doubles, what is the new electric force in terms of F ? (2 marks)
The student’s attempt:
∝ 1 Since F r , F is halved as r doubles. 1 So, the new electric force equals 2 F . © Pearson Education Asia Limited 2015 Can the student get full marks? No. He gets no marks because he states the Coulomb’s law incorrectly. By Coulomb’s law, F ∝ 1 . As r r 2 = 1 = 1 doubles, the new electric force 22 F 4 F . © Pearson Education Asia Limited 2015 Tip 20.2 Electric force
When you use Coulomb’s law to calculate the electric force, make sure you plug the correct numbers. (DSE-13-1B Q11a) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
Two charges of magnitudes 2 μC and 1 μC are separated by a distance of 0.01 m. Find the magnitude of the electric force between them.
1 9 2 −2 Given: = 9 × 10 N m C (2 marks) 4πε0
The student’s attempt:
− − 1 1 (2 × 10 6)(1 × 10 6) = · Qq = · = × −18 F 2 9 2 2.22 10 N 4πε0 r 9 × 10 0.01 © Pearson Education Asia Limited 2015 Can the student get full marks? No. He only gets 1 mark for using Coulomb’s law, but cannot get the mark for the inal answer because he substitutes the numbers wrongly. It is not 4πε but the whole fraction 1 0 4πε0 − that equals 9 × 109 N m2 C 2. The correct answer should be
− − (2 × 10 6)(1 × 10 6) F = 9 × 109 · = 180 N 0.012 © Pearson Education Asia Limited 2015 Tip 20.3 Electric force
When you apply Coulomb’s law to calculate the electric force between two point charges, you do not need to divide the electric force or the separation by two. (DSE-13-1B Q11a) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
Two point charges, each carrying −6 μC, are separated by a distance of 2 mm. Find the magnitude of the electric force between them.
1 9 2 −2 Given: = 9 × 10 N m C (2 marks) 4πε0
The student’s attempt:
− 1 1 1 (6 × 10 6)2 = · · Qq = · × 9 · = F 2 9 10 2 405 N 2 4πε0 r 2 0.02 © Pearson Education Asia Limited 2015 Can the student get full marks? No. He scores no marks as the electric force should not be halved. The correct answer should be − (6 × 10 6)2 F = 9 × 109 · = 810 N 0.022 © Pearson Education Asia Limited 2015 Tip 20.4 Electric force
Note the following points in drawing a free body diagram:
Ï Don’t overlook the weight which is usually hidden in the question.
Ï Never draw a force and its components on the same diagram.
Ï Keep in mind that like charges repel, unlike charges attract, and the electric force always acts along the line joining the centres of two charged spheres.
(DSE-13-1B Q11a) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
A negatively-charged sphere A is hung from the ceiling by a long inextensible string. When another negatively-charged sphere B is brought close to A without touching it, A moves to the position shown and achieve an equilibrium.
Draw the free body diagram of A. (2 marks)
The student’s attempt: © Pearson Education Asia Limited 2015 Can the student get full marks? No. He loses all the marks because
1. he forgets to draw the weight.
2. he draws T and its components on the same free body diagram.
3. The electric force he draws does not act along the line joining the centres of the spheres.
The correct diagram should be as follows. © Pearson Education Asia Limited 2015 Tip 20.5 Electric field
When you are asked to indicate the direction of the resultant electric ield at a point, just draw an arrow:
Ï The direction of the ield at a point is represented by an arrow.
Ï The direction of the arrow should be the same as the direction of the resultant electric force acting on a positive charge.
Ï No need to draw the whole electric ield pattern.
(DSE-13-1B Q11bi) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
Two opposite charges Q and q have the same magnitude. Indicate the direction of the resultant electric ield at point A due to Q and q. (1 mark)
The student’s attempt: © Pearson Education Asia Limited 2015 Can the student get full marks? Yes. The arrow correctly indicates the direction of the resultant electric ield at A, which is pointing to the right. The vertical components of the ields cancel out each other. © Pearson Education Asia Limited 2015 Tip 20.6 Neutral conductor in electric field
Remember that when a neutral conductor is placed near a charged object, they always attract each other, no matter the object is positively or negatively charged. (DSE-13-1B Q11biii) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
As shown, two identical positively-charged spheres X and Y are separated on a horizontal insulating surface. Each of them is attached by an insulated spring which is ixed at one end.
Question: When a neutral metal sphere Z is placed at point A, would the separation between X and Y increase, decrease or remain unchanged? Explain. (2 marks)
The student’s attempt:
The separation between X and Y should remain unchanged, because the sphere Z brought to A carries no net charge, hence exerts no electric force on them. © Pearson Education Asia Limited 2015 Can the student get full marks? No. Quite the opposite, he scores no marks because his answer is simply wrong. The separation between X and Y decreases as the charges induced on the neutral sphere Z produce net attractive forces to both X and Y . © Pearson Education Asia Limited 2015 Chapter 21 Circuit and Power © Pearson Education Asia Limited 2015 Tip 21.1 Circuit
Remember that any open branches in a circuit can be ignored. (DSE-14-1B Q8b) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
In the circuit below, all resistors are identical.
If mode P is selected, which of the following circuits, 1 or 2, shows the correct equivalent circuit? Explain. (2 marks)
The student’s attempt:
Circuit 1, because it consists of four resistors, two in series and two in parallel. © Pearson Education Asia Limited 2015 Can the student get full marks? No. He scores no marks because the equivalent circuit should be circuit 2. To determine the equivalent circuit, we should ignore all the open branches. So, there are only two resistors in series when mode P is chosen. © Pearson Education Asia Limited 2015 Tip 21.2 Circuit
Always read the question carefully and make sure you know what quantities are required to solve in the question. (DSE-12-1B Q8c) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
A resistor R is connected to a 6 V cell as shown below. Its power consumption is 12 W. Calculate the current passing through the resistor R. (2 marks)
The student’s attempt:
V 2 62 R = = = 3 Ω P 12 © Pearson Education Asia Limited 2015 Can the student get full marks? No. He scores no marks because the question asks for the current I instead of the resistance of R. The correct answer should be P 12 I = = = 2 A V 6 © Pearson Education Asia Limited 2015 Tip 21.3 Power
Remember that the power consumed by two resistors connected in parallel is larger than a single resistor under the same voltage. (DSE-12-1B Q8b) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
The igure below shows the structure of a hairdryer using mains supply. It can be operated in low power mode and high power mode, depending on whether the switch S is on or off. When it is in low power mode, the power is 880 W. When it is in high power mode, the power is 2200 W.
Question: What is the resistance of R2? (2 marks)
The student’s attempt:
V 2 2202 R2 = = = 22 Ω P 2200 © Pearson Education Asia Limited 2015 Can the student get full marks? No. He loses 1 mark because he incorrectly substitutes the power consumed in the high power mode in his calculation. To ind the resistance of R2, we should consider the case when the switch is open. This corresponds to the low power mode. So we should substitute 880 W into the equation, instead of 2200 W. The correct answer should be V 2 2202 R2 = = = 55 Ω P 880 © Pearson Education Asia Limited 2015 Tip 21.4 Power
When you explain something, state clearly the relationship between the key variables of the system and the variable concerned in the question. (DSE-14-1B Q8c) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
As shown below, a heating unit consists of a ixed resistor and a rheostat. How should the resistance of the rheostat be set so that the total power dissipation of the heating unit is the largest? Why? (2 marks)
The student’s attempt:
The resistance of the rheostat should be set at maximum. Since the ixed resistor and the rheostat are in parallel, the overall resistance of the resistors is minimum when the resistance of the rheostat is set at maximum. © Pearson Education Asia Limited 2015 Can the student get full marks? No. He loses 1 mark because his answer does not mention how the total power dissipated is affected. To get full marks, the student should further state that: = V 2 By P R , the power dissipation is inversely proportional to the resistance under the same voltage. Therefore, the lowest overall resistance maximizes the total power dissipation. © Pearson Education Asia Limited 2015 Tip 21.5 Internal resistance of voltmeter
When you are asked to explain the effects of the internal resistance of a voltmeter in a circuit, explain it in terms of voltage, not current. (DSE-13-1B Q10b) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
In the circuit shown, two identical resistors are connected in series to a 6 V battery. A voltmeter is connected across XY but the reading is found to be slightly different from the theoretical value of 3 V. Explain why this is so. (2 marks)
The student’s attempt:
The inite internal resistance of the voltmeter lowers the equivalent resistance across XY . Since V ∝ R in series combination, the voltage across XY becomes smaller and a reading slightly smaller than 3 V is expected. © Pearson Education Asia Limited 2015 Can the student get full marks? Yes. As a voltmeter measures voltage, it is much easier for us to explain in terms of voltage instead of current. © Pearson Education Asia Limited 2015 Chapter 22 Ac and Domestic Electricity © Pearson Education Asia Limited 2015 Tip 22.1 Mains electricity
Remember that by V = IR, the current driven by a power source is inversely proportional to the overall resistance of the components connecting to the power source. (DSE-14-1B Q8di) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
In the circuit below, which two of the terminals P, Q and R should be connected to the live wire and the neutral wire so that the current driven from the mains is the largest? (1 mark)
The student’s attempt:
P and Q should be connected to the live wire and the neutral wire. © Pearson Education Asia Limited 2015 Can the student get full marks? No. The current is the largest when the overall resistance is the smallest. Thus P and R should be connected to the live wire and the neutral wire. © Pearson Education Asia Limited 2015 Tip 22.2 Electric shock
Remember that the chance of getting electric shock is higher in wet or humid environment because water provides a conducting path for a current and reduce the resistance between the power source and the human body. (DSE-12-1B Q9a) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
For safety reasons, after washing your hands, you should dry your hands before you touch any electrical device. Explain why.
(2 marks)
The student’s attempt:
The hands get wet after washing. The chance of getting an electric shock is higher. © Pearson Education Asia Limited 2015 Can the student get full marks? No. He only scores 1 mark because his explanation only mentions that the hands are wet. To get full marks, he should also point out the reason why wet hands increase the chances of getting an electric shock. © Pearson Education Asia Limited 2015 Tip 22.3 Installing switch and fuse
Remember that a switch or a fuse should be installed in the live wire. Otherwise the appliance is still ‘live’ even though the appliance is switched off or the fuse is blown. (DSE-14-1B Q8dii) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
Is a switch of an ac appliance installed in the neutral wire as safe as that installed in the live wire? Why? (2 marks)
The student’s attempt:
Yes. It is because no matter which wire the switch is installed, the circuit must be incomplete and no current lows through the appliance when it is switched off. © Pearson Education Asia Limited 2015 Can the student get full marks? No. He scores no marks because he is completely wrong. A switch installed in the neutral wire cannot separate the appliance from high potential. Thus the appliance is still ‘live’ even though it is switched off. © Pearson Education Asia Limited 2015 Chapter 23 Electromagnetism © Pearson Education Asia Limited 2015 Tip 23.1 Charged particle in magnetic field
Remember that the acceleration of a charged particle performing uniform circular motion is always centripetal, i.e. pointing to the centre of its circular path. (DSE-13-1B Q4aii) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
The igure below shows the trajectory of a proton in a uniform magnetic ield. Indicate the directions of the acceleration and the magnetic force acting on the proton at A. (2 marks)
The student’s attempt: © Pearson Education Asia Limited 2015 Can the student get full marks? No. Although he correctly points out the direction of the magnetic force, he is wrong about the direction of the acceleration. So he loses 1 mark. Since the magnetic force provides the centripetal force, the acceleration of the proton points to the centre of the circular path, not in the same direction as the velocity. © Pearson Education Asia Limited 2015 Tip 23.2 Charged particle in magnetic field
When you are asked to explain why the speed of a charged particle remains unchanged in a uniform magnetic ield, you must explain in terms of work and kinetic energy. (DSE-13-1B Q4b) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
Although a magnetic force acts on a charged particle in a uniform magnetic ield, the speed of the charged particle remains unchanged. Why? (2 marks)
The student’s attempt:
The speed remains unchanged because the magnetic force is always perpendicular to the velocity of the particle. © Pearson Education Asia Limited 2015 Can the student get full marks? No. He only scores 1 mark for mentioning that F ⊥ v. Since the magnetic force is always perpendicular to the velocity of the particle, no work is done by the magnetic force and hence the particle cannot gain any kinetic energy. The statement ’No work is done on the particle’ is worth 1 mark. © Pearson Education Asia Limited 2015 Tip 23.3 Charged particle in magnetic field
Make use of ratio when you relate one quantity to another from a known equation. (DSE-13-1B Q4c) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
An electron of mass m carrying a charge −e performs circular motion at a uniform speed v in a uniform magnetic ield B. The radius of its trajectory r is given by mv r = eB
Question: Deduce how the speed of the electron changes if the radius of the trajectory doubles. (2 marks)
The student’s attempt:
The speed of the electron also doubles. © Pearson Education Asia Limited 2015 Can the student get full marks? No. He loses 1 mark because he does = mv not give any explanation. From r eB , the radius r is directly proportional to v. So, v doubles as r doubles. Note that m and e of an electron cannot be varied. © Pearson Education Asia Limited 2015 Chapter 24 Electromagnetic Induction © Pearson Education Asia Limited 2015 Tip 24.1 Induced emf and current
Remember that a changing magnetic ield induces emf, but a constant magnetic ield does not induce any emf. (DSE-14-1B Q9a) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
Refer to the set-up below, state and explain what happens to the galvanometer after the switch is closed.
(4 marks)
The student’s attempt:
After the switch is closed, current lows through A and produces a magnetic ield. The magnetic ield induces an emf in B. So the galvanometer keeps de lecting. © Pearson Education Asia Limited 2015 Can the student get full marks? No. He score no marks because he is completely wrong. An emf is induced in the conductor only when the conductor experiences a change in magnetic ield ( lux). The correct answer should be as follows. At the instant the switch is closed, the current in A increases. Therefore, B experiences a change in magnetic ield. An emf is induced in B and the galvanometer de lects. When the switch is kept closed, the current in A keeps constant. Therefore, B experiences no change in magnetic ield. Thus no emf is induced in B and the galvanometer does not de lect. © Pearson Education Asia Limited 2015 Tip 24.2 Lenz’s law
Remember the aims and the expected results of different demonstrations about eddy current. (DSE-14-1B Q9a) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
Two demonstration set-ups are shown below. Which one is the most suitable to demonstrate Lenz’s law in electromagnetic induction? Which one is the most suitable to demonstrate the braking effect of eddy current?
(1 mark)
The student’s attempt:
Set-up 1 is the most suitable to demonstrate Lenz’s law in electromagnetic induction. Set-up 2 is the most suitable to demonstrate the braking effect of eddy current. © Pearson Education Asia Limited 2015 Can the student get full marks? Yes. In set-up 1, when the dc supply is switched on, the ring jumps and then falls back to the coil. This shows that emf and current are induced against the change of the magnetic lux. In set-up 2, when the magnet is dropped into the aluminium tube, it falls down with a smaller acceleration than g. This shows that the eddy current induced in the tube slows down the magnet. © Pearson Education Asia Limited 2015 Tip 24.3 Lenz’s law
If you are asked to describe ‘what happens / what would be observed’, give a precise answer. Never just say ‘no change’ or ‘no observation’, etc. even the concerned object does not move. (DSE-14-1B Q9bii) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
As shown, a broken metal ring is resting on a horizontal smooth iron bar. Describe what would be observed if the switch is closed. (1 mark)
The student’s attempt:
There is no observation if the switch is closed. © Pearson Education Asia Limited 2015 Can the student get full marks? No. He should say ‘The broken ring does not move.’ This answer is precise enough to score 1 mark. © Pearson Education Asia Limited 2015 Tip 24.4 Transformer
Remember that the live wire is always at high potential. Anyone touching the live wire provides a low resistance path for the current to pass through and therefore gets an electric shock. (DSE-12-1B Q9bi) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
As shown, a transformer is connected to mains supply. If a person touches the live wire in the primary circuit of the transformer, would he get an electric shock? Explain. (2 marks)
The student’s attempt:
The person would not get electric shock because the primary circuit is earthed, and the current in the circuit will low to the ground due to low resistance. So, no current lows through his body. © Pearson Education Asia Limited 2015 Can the student get full marks? No, he scores no marks. Actually, the primary coil converts electrical energy to magnetic energy effectively and therefore has a high ‘effective resistance’. When a person touches the live wire of the primary circuit, his body acts as a short cut to the ground. So, the person will get an electric shock. © Pearson Education Asia Limited 2015 Tip 24.5 Transformer
Remember that no current is induced in the secondary coil unless the secondary circuit of a transformer forms a complete circuit. (DSE-12-1B Q9bii) © Pearson Education Asia Limited 2015
Trap A student a empts the following ques on as shown. Can he get full marks?
The primary coil of a step-up transformer is connected to the mains supply. If a person touches one of the terminals in the secondary coil, will he get an electric shock? Explain. (2 marks)
The student’s attempt:
The person will not get an electric shock because the secondary coil is isolated from high voltage. © Pearson Education Asia Limited 2015 Can the student get full marks? No. Although he correctly states that the person will not get an electric shock, he loses 1 mark for wrong explanation. As the primary coil is connected to an ac supply, an emf is induced in the secondary coil. Therefore, the secondary circuit keeps at high potential and the statement ‘The secondary coil is isolated from high voltage’ is wrong. Actually, the person will not get electric shock because he only touches one terminal of the secondary coil. In this case, his body and the secondary coil do not form a complete circuit. So, no current is induced in his body. This is the correct explanation to the trap above. Note that if a person touches all the two terminals of the secondary coil at the same time, he will get an electric shock because his body and the secondary coil forms a complete circuit now.