QUESTION 8 [16 marks]

Question

QUESTION 8.1
A small isolated sphere A, with a mass of 0,05 kg, carrying an unknown charge
Q, is suspended from a horizontal surface by a string of negligible mass, as
shown in Figure 8 below. A second sphere B, carrying a charge of -6 𝜇C, on an
isolated stand, repels sphere A so that the string forms an angle of 15º to the
vertical. The horizontal distance between the centres of the two spheres is 27
cm. The tension in the string is given as 0,507 N.
Figure 8: Two spheres.
22
Question 8.1.1 [2 marks]
Define Coulomb's Law.
Question 8.1.2 [3 marks]
Draw a VECTOR DIAGRAM of the forces acting on sphere A.
Question 8.1.3 [6 marks]
Calculate the magnitude of the charge on sphere A.
QUESTION 8.2
Two point charges, Q1 and Q2
, a distance of 300 cm apart, are shown below. The
charge on Q1
is -14 μC and the charge on Q2
is +20 μC. See Figure 9 below.
Figure 9: Electric Fields of Linear Charges.
Question 8.2.1 [5 marks]
Calculate the net electric field at point X, which is located 1 m away from Q1 and
2 m away from Q2
.
TOTAL: 16 marks
23
QUESTION 9 [10 marks]
A circular coil with 325 windings (turns) and a diameter of 64 cm, is rotated
clockwise inside a magnetic field with a field strength of 0.13 T. See Figure 10
below:
Figure 10: Rotation of a coil in a magnetic field.
Question 9.1 [3 marks]
Calculate the magnetic flux through the coil at the position indicated on the
diagram, where the coil is perpendicular to the field.
Question 9.2 [4 marks]
If the coil rotates clockwise to 45°, and the potential difference induced is 1,8 V,
calculate the time in which this rotation took place.
24
Question 9.3 [1 mark]
If the circular coil is replaced with a square coil with a side length of 64 cm, and
the same movement is made in the same amount of time, will the induced emf
be the same as, larger than or smaller than the circular coil?
Write down only THE SAME AS, LARGER THAN or SMALLER THAN.
Question 9.4 [2 marks]
Explain the answer to QUESTION 9.3
QUESTION 10 [17 marks]
The circuit below consists of one resistor in series and three resistors connected
in parallel. Consider the circuit diagram shown in Figure 11 below.
Figure 11: Circuit diagram.
Question 10.1 [2 marks]
Define Ohm’s law.
26
Question 10.2 [4 marks]
Calculate the effective resistance of the circuit if the switch is open.
Question 10.3 [3 marks]
Calculate the value of A1 if the switch is open.
Question 10.4 [2 marks]
Calculate the value of V3 if the switch is open.
Question 10.5 [3 marks]
Calculate energy dissipated by the 4Ω resistor in 2 minutes when the switch is
open.
Question 10.6 [3 marks]
How will the value of A1 change if the switch is now closed? Write only
INCREASE, DECREASE, or REMAINS THE SAME. Explain your answer.
.

Answer 1

QUESTION 8

QUESTION 8.1.1 [2 marks]

Define Coulomb's Law.
Coulomb's Law states that the force $F$ between two point charges is directly proportional to the product of the magnitudes of the charges $Q_1$ and $Q_2$, and inversely proportional to the square of the distance $r$ between them. Mathematically, it can be expressed as:
$$F=k\frac{|Q_1{Q}_2|}{r^2}$$
where $k$ is Coulomb's constant ($8.99\times {10}^9,{\text{N m}}^2/{\text{C}}^2$).

QUESTION 8.1.2 [3 marks]

Draw a VECTOR DIAGRAM of the forces acting on sphere A.
(Assuming the drawing capability is unavailable, a description of the diagram will suffice.)

• Draw a vertical arrow pointing downward labeled $W$ representing the gravitational force $W=mg$.
• Draw an upward arrow labeled $T$ for the tension in the string.
• Draw a horizontal arrow pointing to the right labeled $F_e$ for the electric force due to the repulsion from sphere B. The angle between the tension and vertical should be 15°.

QUESTION 8.1.3 [6 marks]

Calculate the magnitude of the charge on sphere A.

1. Resolve the forces acting on sphere A. The vertical components yield:
   $$T\cos ({15}^{\circ })=mg$$ $$0.507\cos ({15}^{\circ })=0.05\cdot 9.81$$

Calculating $T\cos ({15}^{\circ })$:
$$T\cos ({15}^{\circ })=0.507\cdot 0.9659\approx 0.489,\text{N}$$

Calculating the weight $mg$:
$$0.05\cdot 9.81\approx 0.4905,\text{N}$$ This confirms that the vertical forces balance.

2. For the horizontal forces, we can set up the relationship:
   $$T\sin ({15}^{\circ })=F_e$$
   Substituting $T=0.507$:
   $$0.507\sin ({15}^{\circ })=F_e$$ Calculating $F_e$:
   $$F_e\approx 0.507\cdot 0.2588\approx 0.131,\text{N}$$

3. Using Coulomb's law to find $Q$:
   $$F_e=k\frac{|Q\cdot (-6\times {10}^{-6})|}{(0.27{)}^2}$$ Substituting:
   $$0.131=8.99\times {10}^9\frac{|Q\cdot (-6\times {10}^{-6})|}{0.0729}$$

4. Simplifying and calculating $Q$:
   $$|Q|=\frac{0.131\cdot 0.0729}{8.99\times {10}^9\cdot (-6\times {10}^{-6})}$$
   $$|Q|=\frac{0.0095659}{-5.394\times {10}^4}\approx 1.77\times {10}^{-10},\text{C}$$

Therefore, the magnitude of charge on sphere A is approximately:
$$Q\approx 1.77\times {10}^{-10},\text{C}$$

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QUESTION 8.2.1 [5 marks]

Calculate the net electric field at point X.

• The distance from $Q_1$ to point $X$ is 1 m, and the distance from $Q_2$ to $X$ is 2 m.
• The electric field $E$ due to a point charge $Q$ is given by: $$E=k\frac{|Q|}{r^2}$$

1. Electric Field due to $Q_1$:
   $$E_1=k\frac{|Q_1|}{r_1^2}=8.99\times {10}^9\frac{14\times {10}^{-6}}{1^2}=125.86\times {10}^3,\text{N/C}$$

2. Electric Field due to $Q_2$:
   $$E_2=k\frac{|Q_2|}{r_2^2}=8.99\times {10}^9\frac{20\times {10}^{-6}}{2^2}=44.95\times {10}^3,\text{N/C}$$

3. Direction of E-fields:

• $E_1$ points towards $Q_1$ (attractive, as $Q_1$ is negative).
• $E_2$ points away from $Q_2$ (repulsive, as $Q_2$ is positive).

4. Net Electric Field at Point X:
   $$E_{net}=E_1-E_2=125.86\times {10}^3-44.95\times {10}^3$$ $$E_{net}=80.91\times {10}^3,\text{N/C}$$

Thus, the net electric field at point X is approximately:
$$E_{net}\approx 80.91\times {10}^3,\text{N/C}$$

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QUESTION 9

QUESTION 9.1 [3 marks]

Calculate the magnetic flux through the coil at the position indicated on the diagram, where the coil is perpendicular to the field.

• The area $A$ of the circular coil is given by: $$A=\pi r^2$$

Calculating $r$: $$r=\frac{64,\text{cm}}{2}=0.32,\text{m}$$

Calculating the area: $$A=\pi (0.32{)}^2\approx 0.3217,{\text{m}}^2$$

Magnetic flux $\mathrm{\Phi }$ is given by: $$\mathrm{\Phi }=BA$$ $$\mathrm{\Phi }=0.13\times 0.3217\approx 0.0418,\text{Wb}$$

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QUESTION 9.2 [4 marks]

If the coil rotates clockwise to 45°, and the potential difference induced is 1.8 V, calculate the time in which this rotation took place.
We use Faraday's law of induction: $$\text{emf}=-\frac{d\mathrm{\Phi }}{dt}$$

1. Find the initial flux ${\mathrm{\Phi }}_i$: $${\mathrm{\Phi }}_i=BA\cos (0^{\circ })=0.13\cdot 0.3217=0.0418,\text{Wb}$$

2. Find the final flux ${\mathrm{\Phi }}_f$: $${\mathrm{\Phi }}_f=BA\cos ({45}^{\circ })=0.13\cdot 0.3217\cdot \frac{1}{\sqrt{2}}\approx 0.0295,\text{Wb}$$

3. Change in flux $\mathrm{\Delta }\mathrm{\Phi }$: $$\mathrm{\Delta }\mathrm{\Phi }={\mathrm{\Phi }}_f-{\mathrm{\Phi }}_i\approx 0.0295-0.0418\approx -0.0123,\text{Wb}$$

4. Now calculate the induced emf: $$1.8=|-\frac{-0.0123}{dt}|$$ $$dt=\frac{0.0123}{1.8}\approx 0.00683,\text{s}$$

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QUESTION 9.3 [1 mark]

If the circular coil is replaced with a square coil, will the induced emf be the same as, larger than or smaller than the circular coil?
Answer: SMALLER THAN.

QUESTION 9.4 [2 marks]

Explain the answer to QUESTION 9.3
The induced emf depends on the area of the coil and the rate of change of flux. The square coil has a smaller area compared to the circular coil (with the same side length or diameter) when considering rotation in the same magnetic field. Therefore, the induced emf in the square coil will be smaller than that in the circular coil due to the geometrical differences impacting the effective area in the magnetic field.

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QUESTION 10

QUESTION 10.1 [2 marks]

Define Ohm’s law.
Ohm's Law states that the current $I$ flowing through a conductor between two points is directly proportional to the voltage $V$ across the two points and inversely proportional to the resistance $R$ of the conductor. Mathematically, it is expressed as:
$$I=\frac{V}{R}$$

QUESTION 10.2 [4 marks]

Calculate the effective resistance of the circuit if the switch is open.
Assuming $R_1=4\mathrm{\Omega }$ (series) and $R_2,R_3,R_4$ are in parallel. Let's denote $R_p$ as the equivalent resistance of the parallel section:

1. Calculate $R_p$: $$\frac{1}{R_p}=\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$$

2. Assume $R_2=R_3=R_4=8\mathrm{\Omega }$: $$\frac{1}{R_p}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$$ $$R_p=\frac{8}{3}\approx 2.67,\mathrm{\Omega }$$

3. Total effective resistance $R_{total}$: $$R_{total}=R_1+R_p=4+2.67\approx 6.67,\mathrm{\Omega }$$

QUESTION 10.3 [3 marks]

Calculate the value of A1 if the switch is open.
Using Ohm's Law and knowing the total voltage $V$ (assume given $V=12V$): $$A_1=\frac{V}{R_{total}}=\frac{12}{6.67}\approx 1.80,\text{A}$$

QUESTION 10.4 [2 marks]

Calculate the value of V3 if the switch is open.
Using Ohm's Law for the resistor $R_1=4\mathrm{\Omega }$: $$V_3=I\cdot R_1=1.80\cdot 4\approx 7.20,\text{V}$$

QUESTION 10.5 [3 marks]

Calculate energy dissipated by the 4Ω resistor in 2 minutes when the switch is open.
Energy $E=P\cdot t$ where $P=I^{2}R$:

1. Calculate power: $$P=(1.80{)}^2\times 4\approx 12.96,\text{W}$$
2. Convert time to seconds: $t=2\times 60=120,\text{s}$: $$E=12.96\cdot 120\approx 1555.2,\text{J}$$

QUESTION 10.6 [3 marks]

How will the value of A1 change if the switch is now closed?
Answer: INCREASE.
Explanation: Closing the switch reduces the total resistance in the circuit because the parallel branches present a lower equivalent resistance compared to when the switch is open. According to Ohm's Law, if resistance decreases and the voltage remains constant, the current $I$ (and thus $A_1$) will increase.