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.
- 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.
-
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 \( F_e \):
\[ F_e \approx 0.507 \cdot 0.2588 \approx 0.131 , \text{N} \] -
Using Coulomb's law to find \( Q \):
\[ F_e = k \frac{|Q \cdot (-6 \times 10^{-6})|}{(0.27)^2} \] Substituting:
\[ 0.131 = 8.99 \times 10^9 \frac{|Q \cdot (-6 \times 10^{-6})|}{0.0729} \] -
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:
\[ Q \approx 1.77 \times 10^{-10} , \text{C} \]
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: \[ E = k \frac{|Q|}{r^2} \]
-
Electric Field due to \( Q_1 \):
\[ E_1 = k \frac{|Q_1|}{r_1^2} = 8.99 \times 10^9 \frac{14 \times 10^{-6}}{1^2} = 125.86 \times 10^3 , \text{N/C} \] -
Electric Field due to \( Q_2 \):
\[ E_2 = k \frac{|Q_2|}{r_2^2} = 8.99 \times 10^9 \frac{20 \times 10^{-6}}{2^2} = 44.95 \times 10^3 , \text{N/C} \] -
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).
- 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:
\[ E_{net} \approx 80.91 \times 10^3 , \text{N/C} \]
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 \( \Phi \) is given by: \[ \Phi = B A \] \[ \Phi = 0.13 \times 0.3217 \approx 0.0418 , \text{Wb} \]
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:
\[ \text{emf} = -\frac{d\Phi}{dt} \]
-
Find the initial flux \( \Phi_i \): \[ \Phi_i = B A \cos(0^\circ) = 0.13 \cdot 0.3217 = 0.0418 , \text{Wb} \]
-
Find the final flux \( \Phi_f \): \[ \Phi_f = B A \cos(45^\circ) = 0.13 \cdot 0.3217 \cdot \frac{1}{\sqrt{2}} \approx 0.0295 , \text{Wb} \]
-
Change in flux \( \Delta \Phi \): \[ \Delta \Phi = \Phi_f - \Phi_i \approx 0.0295 - 0.0418 \approx -0.0123 , \text{Wb} \]
-
Now calculate the induced emf: \[ 1.8 = \left| -\frac{-0.0123}{dt} \right| \] \[ dt = \frac{0.0123}{1.8} \approx 0.00683 , \text{s} \]
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.
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 \( R_1 = 4 \Omega \) (series) and \( R_2, R_3, R_4 \) are in parallel. Let's denote \( R_p \) as the equivalent resistance of the parallel section:
-
Calculate \( R_p \): \[ \frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} \]
-
Assume \( R_2 = R_3 = R_4 = 8 \Omega \): \[ \frac{1}{R_p} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8} \] \[ R_p = \frac{8}{3} \approx 2.67 , \Omega \]
-
Total effective resistance \( R_{total} \): \[ R_{total} = R_1 + R_p = 4 + 2.67 \approx 6.67 , \Omega \]
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 \( R_1 = 4 \Omega \):
\[ V_3 = I \cdot R_1 = 1.80 \cdot 4 \approx 7.20 , \text{V} \]
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 \( P = I^2 R \):
- Calculate power: \[ P = (1.80)^2 \times 4 \approx 12.96 , \text{W} \]
- 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.