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Transcript
International Indian School, Riyadh, K.S.A. Mohammad Abdul Mumeed, Senior Secondary Teacher in Physics Electric Forces & Electric Fields(With Reasoning & Solution) These questions contain both Concepts & Calculations and are designed to enhance the subject knowledge of the students. Advised to work alone or in a small group before checking the Solution i.e. for Horizontal- learning Numerical Question#1: Two very small spheres are initially neutral and separated by a distance of 0.50m. Suppose that 3.0x10 13 electrons are removed from one sphere and placed on the other. a) What is the magnitude of the electrostatic force that acts on each sphere? b) Is the force attractive or repulsive? Why? REASONING AND SOLUTION Initially, the two spheres are neutral. Since negative charge is removed from the sphere which loses electrons, it then carries a net positive charge. Furthermore, the neutral sphere to which the electrons are added is then negatively charged. Once the charge is transferred, there exists an electrostatic force on each of the two spheres, the magnitude of which is given by Coulomb's law SOLUTION Since each electron carries a charge of the amount of negative charge removed from the first sphere is Thus, the first sphere carries a charge while the second sphere carries a charge 1|Page International Indian School, Riyadh, K.S.A. Mohammad Abdul Mumeed, Senior Secondary Teacher in Physics The magnitude of the electrostatic force that acts on each sphere is, therefore, b. Since the spheres carry charges of opposite sign, the force is attractive. Numerical Question#2: A tiny ball of mass 0.012 kg carries a charge of -18 µC. What electric field (magnitude and direction) is needed to cause the ball to float above the ground? REASONING AND SOLUTION Two forces act on the charged ball (charge q); they are the downward force of gravity mg and the electric force F due to the presence of the charge q in the electric field E. In order for the ball to float, these two forces must be equal in magnitude and opposite in direction, so that the net force on the ball is zero (Newton's second law). Therefore, F must point upward, which we will take as the positive direction. According to Equation F = qE. Since the charge q is negative, the electric field E must point downward, as the product qE in the expression F = qE must be positive, since the force F points upward. The magnitudes of the two forces must be equal, so that . This expression can be solved for E. SOLUTION The magnitude of the electric field E is As discussed in the reasoning, this electric field points downward. 2|Page