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Solutions 2011 Question 9 (a) (i) State Coulomb’s law. The force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. (ii) What is the new force, in terms of F , between the spherical conductors?
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(b) (i) Draw a labelled diagram of an electroscope.
A = insulated joint, B = metal case (ii) Why should the frame of an electroscope be earthed? If the frame was charged it would affect the degree of deflection of the leaf. (iii) Describe how to charge an electroscope by induction.
- Bring a charged rod near the electroscope (the positive and negative charges become separated on it).
- Keeping the charged rod in place, earth the cap by touching it with your finger. Some of the negative charge on the metal flows through you to earth.
- Remove your finger, then and only then remove the rod.
- The conductor will now be positively charged.
(c) (i) How does a full-body metal-foil suit protect an operator when working on high voltage power lines? All charges will reside on the outside of the conducting suit (because the suit blocks out external electrical fields) so he won’t get shocked. (ii) Describe an experiment to investigate the principle by which the operator is protected.
- Charge the conductor (a metal can will do fine).
- Using a proof plane, touch the inside of the can and bring it up to the GLE. Notice that there is no deflection.
- Touch the proof plane off the outside of the can and bring it up to the GLE. Notice that there is a deflection.
- Conclusion: charge resides on outside only
2010 Question 12 (d) (i) Define electric field strength and give its unit of measurement. Electric field strength is defined as force per unit charge. Its unit is the N C– (ii) Copy the diagram into your answerbook and show on it the direction of the electric field at point P.
(iii) Calculate the electric field strength at P. The electric field strength at P is the sum of the electric fields acting on P from the other two charges. The electric field strength is towards the left in both cases (attracted to the negative charge and repelled from the positive charge). Because they are both in the same direction the individual field strengths can simply be added together.
2 × 10−
4 𝜋 ε(0.1)^2
5 × 10−
4 𝜋 ε(0.15)^2
Etotal = 3.77 × 10^6 N C-
(iv) Under what circumstances will point discharge occur? Large electric field strength /potential at a point / high charge density at a point.
2009 Question 9 (i) Define potential difference Potential difference is the work done in moving unit charge from one place to another. (ii) Define capacitance The capacitance of a conductor is the ratio of the charge on the conductor to its potential. (iii) A capacitor stores energy. Describe an experiment to demonstrate that a capacitor stores energy.
- Set up as shown.
- Close the switch to charge the capacitor.
- Remove the battery and connect the terminals together to ‘short’ the circuit.
- The bulb will flash as the capacitor discharges, showing that it stores energy.
The ability of a capacitor to store energy is the basis of a defibrillator. During a heart attack the chambers of the heart fail to pump blood because their muscle fibres contract and relax randomly. To save the victim, the heart muscle must be shocked to re-establish its normal rhythm. A defibrillator is used to shock the heart muscle. A 64 μF capacitor in a defibrillator is charged to a potential difference of 2500 V. The capacitor is discharged through electrodes attached to the chest of a heart attack victim. (iv) Calculate the charge stored on each plate of the capacitor. q = CV q = (64 × 10-6)(2500) q = 0.16 C (v) Calculate the energy stored in the capacitor. E = ½ CV^2 = ½ (64 × 10-6)(2500)^2 = 200 J (vi) Calculate the average current that flows through the victim when the capacitor discharges in a time of 10 ms. I = q/t = (64 × 10-6)/(10 × 10-3) = 16 A (vii) Calculate the average power generated as the capacitor discharges. P = W/t = (200)/(10 × 10-3) = 20000 W
2006 Question 12 (b) (i) List the factors that affect the capacitance of a parallel plate capacitor. Common area of plates, distance apart, permittivity of dielectric between plates. (ii) The plates of an air filled parallel plate capacitor have a common area of 40 cm^2 and are 1 cm apart. The capacitor is connected to a 12 V d.c. supply. Calculate the capacitance of the capacitor. C= ε A/d C = [(8.85 × 10-12)(40 × 10-4)] / (0.01) C = 3.54 × 10-12^ F (iii) Calculate the magnitude of the charge on each plate. Q = C V Q = (3.54 x 10-12)(12) = 4.2(5) x 10-11^ C (iv) What is the net charge on the capacitor? zero (v) Give a use for a capacitor. blocks d.c. /smoothing /tuning circuits / timing circuits / flash guns for cameras.
2005 Question 10 (i) Define electric field strength****. Electric field strength is defined as force per unit charge. (ii) State Coulomb’s law of force between electric charges. The force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. (iii) Why is Coulomb’s law an example of an inverse square law? Force is inversely proportional to distance squared. (iv) Give two differences between the gravitational force and the electrostatic force between two electrons. Gravitational force is much smaller than the electrostatic force. Gravitational force is attractive, electrostatic force (between two electrons) is repulsive. (v) Describe an experiment to show an electric field pattern. High voltage and two metal plates /electrodes Semolina and oil in container Connect a (high) voltage to the plates in container Semolina lines up in the field (vi) Calculate the electric field strength at the point B, which is 10 mm from an electron. E = Q/4πεd^2 = (1.6 × 10-19)/4π(8.9 × 10-12)(0.01)^2 E = 1.4 × 10-5^ N C- (vii) What is the direction of the electric field strength at B? Towards the electron / to the right (viii) A charge of 5 μC is placed at B. Calculate the electrostatic force exerted on this charge. F = Eq or F = (1.4 × 10-5)(5 × 10-6) = 7.2 × 10-11^ N Towards the electron
2004 Question 8 (i) Define potential difference. The Potential difference (p.d.) between two points is the work done in bringing a charge of 1 Coulomb from one point to the other. (ii) Define capacitance. The Capacitance of a conductor is the ratio of the charge on the conductor to its potential. (iii) Describe an experiment to demonstrate that a capacitor can store energy.
- Set up as shown.
- Close the switch to charge the capacitor.
- Remove the battery and connect the terminals together to ‘short’ the circuit.
- The bulb will flash as the capacitor discharges, showing that it stores energy.
(iv) Calculate the potential difference across the resistor and hence the potential difference across the capacitor when the current is 80 μA. V = IR V (across 47 kΩ resistor) = (80 ×10−^6 )(47 ×10^3 ) = 3.76 V V (across the capacitor) = 6 − 3.76 = 2.24 V (v) Calculate the charge on the capacitor at this instant.
C = Q/V Q = CV = (50 ×10−^6 )(2.24) = 1.12 × 10-4^ C
(vi) Calculate the energy stored in the capacitor when it is fully charged. E = ½ CV^2 = ½ (50 ×10−^6 )(6 )^2 = 9 ×10−^4 J (vii) Describe what happens in the circuit when the 6 V d.c. supply is replaced with a 6 V a.c. supply. The current will flow continually.
2003 Question 12 (c) (i) State Coulomb’s law of force between electric charges. Coulomb’s Law states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. (ii) Define electric field strength and give its unit. Electric field strength at a point is the force per unit charge at that point. The unit of electric field strength is the Newton per Coulomb (NC-1). (iii) How would you demonstrate an electric field pattern? Oil and semolina or seeds High tension / high voltage Lines of semolina show field (iv) The diagram shows a negative charge – Q at a point X. Copy the diagram and show on it the direction of the electric field strength at Y. Arrow towards X
