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Three point charges of +2 uC , -3uC , and -3uC are kept at the vertices A,B and C respectively of an equilateral triangle of side 20 cm as shown in the figure(attached with question).What should be the sign and magnitude of the charge to be placed at the midpoint M of the side BC so that the charge at A remain in equilibruium. (Delhi -2005) |
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Answer» gram.Let K = 1/(4πε) = 9 * 10⁹ N-m²/C²The forces F1 and F2 on the charge at Q are along the sides AB and AC. magnitude of F1 = F2 = K * 2 * 10⁻⁶ * 3 * 10⁻⁶ / 0.20² = 1.5 * 10⁻⁸ K NewtonsTheir COMPONENTS along the horizontal directions are equal to F1 COS 60° and F2 Cos 60° which are equal in magnitude and opposite in direction. They cancel.Their components along AM will add to : 2 F1 Cos 30° = 2 * 1.5 * 10⁻⁸ * √3/2 * K newtons = 1.5√3 *10⁻⁸ K NewtonsLet the charge placed at M be q Coulombs. AM² = 0.20² - 0.10² = 0.03 m²Force due to q on charge at A will be along MA: F3 F3 = K q * 2 * 10⁻⁶ / 0.03 = 6.667 q K * 10⁻⁹ Newtons For the charge at A to be stationary, 6.667 q K * 10⁻⁹ = 1.5 √3 * 10⁻⁸ K q = 3.897 Coulombs positive charge. |
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