(a) Define:(i) the coulomb; (ii) resistance. (b) State the three effects of an electric current. (c) State the standard international colour convention for ...

(a)(i) The coulomb is the quantity of electric charge that passes a point in a circuit when a steady current of one ampere flows for one second: \( 1\ \text{C} = 1\ \text{A}\times1\ \text{s} \).

(a)(ii) Resistance is the opposition offered by a conductor to the flow of electric current through it. It is the ratio of the potential difference across the conductor to the current through it, \( R = \dfrac{V}{I} \), measured in ohms \( (\Omega) \).

(b) Three effects of an electric current:

• Heating (thermal) effect.
• Magnetic effect.
• Chemical effect (electrolysis).

(c) Standard international colour convention:

• (i) Live: brown.
• (ii) Neutral: blue.
• (iii) Earth: green-and-yellow (striped).

(d) From the diagram, the fulcrum divides the uniform rod so that the \( 5.0\ \text{kg} \) watermelon hangs at the left end, a distance \( L_1 = 1.0\ \text{m} \) to the left of the pivot. The rod is \( 4.0\ \text{m} \) long and uniform, so its weight (\( 4.0\ \text{kg} \)) acts at its centre, which is \( L_2 = 1.0\ \text{m} \) to the right of the pivot. The right end carries \( +q \), a distance \( 3.0\ \text{m} \) to the right of the pivot, with \( -q \) fixed \( 2.5\ \text{m} \) directly below it. Take \( g = 10\ \text{m s}^{-2} \).

(i) Anti-clockwise moment (produced by the watermelon at the left end):

\[ W_{\text{melon}} = 5.0\times10 = 50\ \text{N} \]\[ \tau_{\text{acw}} = 50\times1.0 = 50\ \text{N m} \]

(ii) Value of \( q \): the clockwise moments are produced by the rod's weight and by the attractive electrostatic force \( F \) which pulls the right end downward.

\[ W_{\text{rod}} = 4.0\times10 = 40\ \text{N},\qquad \tau_{\text{rod}} = 40\times1.0 = 40\ \text{N m} \]

Taking moments about the pivot (anti-clockwise = clockwise):

\[ 50 = 40 + F\times3.0 \]\[ F = \frac{50-40}{3.0} = \frac{10}{3.0} = 3.33\ \text{N} \]

This is the Coulomb force between the two charges \( 2.5\ \text{m} \) apart:

\[ F = \frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{r^{2}} = \frac{9\times10^{9}\,q^{2}}{(2.5)^{2}} \]\[ 3.33 = \frac{9\times10^{9}\,q^{2}}{6.25} \]\[ q^{2} = \frac{3.33\times6.25}{9\times10^{9}} = \frac{20.8}{9\times10^{9}} = 2.31\times10^{-9} \]\[ q = \sqrt{2.31\times10^{-9}} = 4.8\times10^{-5}\ \text{C} \]

Therefore \( q \approx 4.8\times10^{-5}\ \text{C}\ (48\ \mu\text{C}) \).