10 Charges of \( +2.6 \mu \mathrm{C} \) and \( +1,09 \mu \mathrm{C} \) are a distance of 2.3 cm apart. 10.1 Calculate the electrostatic force between the charges. 10.2 The charges are now moved to touch each other. 10.3 If the system is earthed, how many electrons will flow in/out the system? 10.4 If the system? what will the charge of each object be now?

To calculate the electrostatic force between the two charges, you can use Coulomb's law, which states: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( F \) is the magnitude of the force, \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them in meters. For your charges, \( q_1 = +2.6 \mu \mathrm{C} = 2.6 \times 10^{-6} \, \mathrm{C} \), \( q_2 = +1.09 \mu \mathrm{C} = 1.09 \times 10^{-6} \, \mathrm{C} \), and the distance \( r = 2.3 \, \text{cm} = 0.023 \, \text{m} \). Plugging the values into Coulomb's law: \[ F = 8.99 \times 10^9 \frac{(2.6 \times 10^{-6})(1.09 \times 10^{-6})}{(0.023)^2} \approx 5.10 \, \text{N} \] Now, regarding the charges touching each other, when two conductors touch, they share their charge equally. The total charge \( Q \) is the sum of the individual charges: \[ Q = q_1 + q_2 = 2.6 \mu C + 1.09 \mu C = 3.69 \mu C \] Since both charges are the same (equal sized conductors), the new charge on each will be: \[ \text{Charge on each after touching} = \frac{Q}{2} = \frac{3.69 \mu C}{2} = 1.845 \mu C \] Now, if the system is earthed, electrons may flow to neutralize the excess voltage relative to the ground. To determine how many electrons will flow in or out the system, we first need to find out how many electrons correspond to the charge difference created after earthing. The charge of an electron is approximately \( -1.6 \times 10^{-19} \, \text{C} \). If the system was originally neutral after touching and has a charge of \(1.845 \mu \mathrm{C}\), then electrons will flow in to neutralize this charge. The number of electrons \( n \) that corresponds to this charge is given by: \[ n = \frac{|Q|}{|e|} = \frac{1.845 \times 10^{-6}}{1.6 \times 10^{-19}} \approx 1.15 \times 10^{13} \] So, approximately \( 1.15 \times 10^{13} \) electrons will flow into the system to neutralize the charge once it is earthed. After earthing, the charge of each object would be \( 0 \mu \mathrm{C} \), as both have been neutralized by the incoming electrons.