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9. Find the force between two small charges in the following cases: a) \( -1.0 \times 10^{-7} \mathrm{C} \) and \( +2.0 \times 10^{-7} \mathrm{C} \) b) \( +4.0 \times 10^{7} \mathrm{C} \) and \( +3.0 \times 10^{-7} \mathrm{C} \) placed 4 cm apart in vacuum. Find in each case the intensity at P midway of the two charges. \( \left[\left(4 \pi \varepsilon_{0}\right)^{-1}=9.0 \times 10^{\circ} \mathrm{N} \mathrm{m}^{2} \mathrm{C}^{-2}\right] \)

Ask by King May. in Ghana
Mar 15,2025

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- **Case (a)**: - Force: \( -7.70625 \times 10^{14} \, \mathrm{N} \) (attractive) - Electric Field Intensity at P: \( 2.25 \times 10^{14} \, \mathrm{N/C} \) - **Case (b)**: - Force: \( -9.95625 \times 10^{14} \, \mathrm{N} \) (repulsive) - Electric Field Intensity at P: \( 1.26 \times 10^{15} \, \mathrm{N/C} \)

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To find the force between two charges, we can use Coulomb's law, which states that the electrostatic force \( F \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( k = \frac{1}{4 \pi \varepsilon_0} \). ### a) For charges \( -1.0 \times 10^{-7} \mathrm{C} \) and \( +2.0 \times 10^{-7} \mathrm{C} \): Given: - \( q_1 = -1.0 \times 10^{-7} \mathrm{C} \) - \( q_2 = +2.0 \times 10^{-7} \mathrm{C} \) - \( r = 4 \mathrm{cm} = 0.04 \mathrm{m} \) Using Coulomb's law: \[ F = (9 \times 10^9 \mathrm{N m}^2/\mathrm{C}^2) \frac{|(-1.0 \times 10^{-7})(2.0 \times 10^{-7})|}{(0.04)^2} \] \[ F = 9 \times 10^9 \frac{2.0 \times 10^{-14}}{0.0016} = 1.125 \, \mathrm{N} \] ### b) For charges \( +4.0 \times 10^{-7} \mathrm{C} \) and \( +3.0 \times 10^{-7} \mathrm{C} \): Given: - \( q_1 = +4.0 \times 10^{-7} \mathrm{C} \) - \( q_2 = +3.0 \times 10^{-7} \mathrm{C} \) - \( r = 4 \mathrm{cm} = 0.04 \mathrm{m} \) Using Coulomb's law: \[ F = (9 \times 10^9 \mathrm{N m}^2/\mathrm{C}^2) \frac{|(4.0 \times 10^{-7})(3.0 \times 10^{-7})|}{(0.04)^2} \] \[ F = 9 \times 10^9 \frac{1.2 \times 10^{-13}}{0.0016} = 6.75 \, \mathrm{N} \] ### Intensity at Point P: Point P is midway, so the distance from each charge to P is \( 0.02 \mathrm{m} \). 1. **For the first pair:** \[ E_1 = \frac{F}{q} = \frac{1.125}{2.0 \times 10^{-7}} = 5625 \, \mathrm{N/C} \] The direction is towards the positive charge due to the attractive force. 2. **For the second pair:** \[ E_2 = \frac{F}{q} = \frac{6.75}{3.0 \times 10^{-7}} = 22500 \, \mathrm{N/C} \] The direction is away from the charges as both are positive. Thus, the total intensity at P is \( E_{net} = E_1 - E_2 \), summing the effects of both forces based on their directions.

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