InterviewSolution
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InterviewSolution
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Two fixed charges A and B of 5 pC each are separated by adistance of 6 m. C is the mid point of the line joining A andB. A charge 'Q of -5C is shot perpendicular to the linejoining A and B through C with a kinetic energy of 0.06 J.The charge 'Q comes to rest at a point D. The distance CD[KCET 2012]IS |
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Answer» the charge q = - 5μC starts with a KE of 0.06 J from point C and stops at point d. Initial total energy of the charge q = KE + PE = 0.06 + 9*10⁹ * 5 *10⁻⁶ * (-5 *10⁻⁶)/3 + 9*10⁹ * 5*10⁻⁶*(-5*10⁻⁶)/3 J = 0.06 - 0.075 - 0.075 J = - 0.09 J Let ad = bd = x m Final energy = PE (as KE = 0) = - 2 * 9*10⁹ * 5*10⁻⁶ * 5 * 10⁻⁶ /x J = - 0.450/x J Hence, using Energy conservation we get: 0.450/x = 0.09 x = 5 mcd =√(x² - 3²) = 4 m |
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