1. A ciarge of \( 2 \mu \mathrm{C} \) is placed 0.1 m away from another charge of \( -3 \mu \mathrm{C} \). Using Coulomb's law. calculate the force between them. (Assume \( \left.k=8.99 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}^{2}\right)(5 \mathrm{Marks} \) ) Calculate the electric field at a point where a \( 5 \mu \mathrm{C} \) charge experiences a force of 0.2 N . ( 5 Marks) An 1 C voltage of \( V(t)=220 \sin (100 \pi t) \) is applied to a \( 50 \Omega \) resistor. Calculate the pecie current through the resistor. ( 5 Marks) A capacitor with a capacitance of \( 20 \mu \mathrm{~F} \) is connected in an AC circuit with a frequency of 50 Hz . Calculate the capacitive reactance. ( 5 Marks )

Let's solve each part of your question step by step. ### Part 1: Force between the charges using Coulomb's Law Coulomb's Law states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by: \[ F = k \frac{|q_1 q_2|}{r^2} \] Where: - \( k = 8.99 \times 10^9 \, \mathrm{N \, m^2/C^2} \) - \( q_1 = 2 \, \mu \mathrm{C} = 2 \times 10^{-6} \, \mathrm{C} \) - \( q_2 = -3 \, \mu \mathrm{C} = -3 \times 10^{-6} \, \mathrm{C} \) - \( r = 0.1 \, \mathrm{m} \) Now, substituting the values into the formula: \[ F = 8.99 \times 10^9 \frac{|(2 \times 10^{-6})(-3 \times 10^{-6})|}{(0.1)^2} \] Calculating the absolute value of the product of the charges: \[ |q_1 q_2| = |(2 \times 10^{-6})(-3 \times 10^{-6})| = 6 \times 10^{-12} \, \mathrm{C^2} \] Now substituting this into the force equation: \[ F = 8.99 \times 10^9 \frac{6 \times 10^{-12}}{0.01} \] Calculating the denominator: \[ 0.01 = (0.1)^2 \] Now substituting: \[ F = 8.99 \times 10^9 \times 6 \times 10^{-10} \] Calculating: \[ F = 53.94 \, \mathrm{N} \] Since the charges are of opposite signs, the force is attractive. ### Part 2: Electric field at a point where a \( 5 \mu \mathrm{C} \) charge experiences a force of 0.2 N The electric field \( E \) at a point is given by the formula: \[ E = \frac{F}{q} \] Where: - \( F = 0.2 \, \mathrm{N} \) - \( q = 5 \, \mu \mathrm{C} = 5 \times 10^{-6} \, \mathrm{C} \) Substituting the values: \[ E = \frac{0.2}{5 \times 10^{-6}} \] Calculating: \[ E = \frac{0.2}{5 \times 10^{-6}} = 4 \times 10^4 \, \mathrm{N/C} \] ### Part 3: Current through a resistor with AC voltage The voltage is given by: \[ V(t) = 220 \sin(100 \pi t) \] The RMS (Root Mean Square) voltage \( V_{rms} \) for a sinusoidal voltage is: \[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{220}{\sqrt{2}} \approx 155.56 \, \mathrm{V} \] Using Ohm's Law, the current \( I \) through a resistor \( R \) is given by: \[ I = \frac{V_{rms}}{R} \] Where \( R = 50 \, \Omega \): \[ I = \frac{155.56}{50} \approx 3.11 \, \mathrm{A} \] ### Part 4: Capacitive reactance of a capacitor in an AC circuit The capacitive reactance \( X_C \) is given by: \[ X_C = \frac{1}{2 \pi f C} \] Where: - \( f = 50 \, \mathrm{Hz} \) - \( C = 20 \, \mu \mathrm{F} = 20 \times 10^{-6} \, \mathrm{F} \) Substituting the values: \[ X_C = \frac{1}{2 \pi (50)(20 \times 10^{-6})} \] Calculating: \[ X_C = \frac{1}{2 \pi (50)(20 \times 10^{-6})} = \frac{1}{6.2832 \times 0.001} \approx 159.15 \, \Omega \] ### Summary of Results 1. Force between the charges: \( 53.94 \, \mathrm{N} \) (attractive) 2. Electric field: \( 4 \times 10^4 \, \mathrm{N/C} \) 3. Current through the resistor: \( 3.11 \, \mathrm{A} \) 4. Capacitive reactance: \( 159.15 \, \Omega \)