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Charges of \( +26 \mu \mathrm{C} \) and \( 1.09 \mu \mathrm{C} \) are o distance of 2.3 cm apart. 10.1 Calculate the electrostatic force be wen the charges. \( \qquad \) \( \qquad \) \( \qquad \) 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? \( \qquad \) \( \qquad \) ....................................... placed back in their original positions, 10.4 If the charges are not eorthed object be now? \( \qquad \) \( \qquad \) \( \qquad \) 10.5 Calculate how many electrons moved from the one charge to the other. \( \qquad \) \( \qquad \) \( \qquad \) \( \qquad \)

Ask by Hills Smith. in South Africa
Mar 16,2025

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The electrostatic force between the charges is approximately \( 2.115 \times 10^{16} \, \mathrm{N} \). When the charges touch, they share the total charge of \( 27.09 \, \mu \mathrm{C} \) equally, resulting in each charge becoming \( 13.545 \, \mu \mathrm{C} \). When the system is earthed, about \( 4.62 \times 10^{7} \) electrons flow in or out to neutralize the charge. If the charges are not earthed, they retain the total charge of \( 27.09 \, \mu \mathrm{C} \). Approximately \( 1.90 \times 10^{7} \) electrons moved from one charge to the other when they touched.

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