1.DrFrank NStein was working in his lab late one night trying to charge a large hollow 81.0g aluminum sphere using a Van de Graf generatorHe calculated that he needed to increase the number of electrons on the sphere by 1 x 10-10%How many electrons is he going to need to move to the sphere?2.It turns out that DrStein is somewhat evilHe has created a monster with a human body and a 1000.0 g metal robot head to destroy the world, or at least diminish its self esteemThe robot’s head has an induced charge of +5.63 x 10-6C and is .500m for the sphereA.) What is the force (include direction) between the head and the sphere? 4.The Drtouches the head of the monster with the charged sphere and moves it back to a distance of 1.00mWhat is the force (include direction) now between the head and sphere?
Ionic compounds balance based on their components' ionic chargesThey get those charges based on the number of electrons vsprotons each atom hasIf an atom has 3 protons and 5 electrons, then theoretically it would have an ionic charge of -2, because electrons have a negative charge and protons have a positive chargeWith more electrons than protons, the atom would have a negative charge, and vice versaOxygen has an ionic charge of -2, and Aluminum has an ionic charge of +3Given the equation Al2O3, there are 2 Aluminum atoms and 3 Oxygen atomsTherefore: 2 Aluminum (+3 2 +6) 3 Oxygen ( -2 3 -6) When you add +6 to -6, you get 0, which is in balanceHope this helps!
1We may suppose that the ratio given refers to the number of electrons in the aluminum of the sphere, which is three molesThat number would be 3 x 27 x 6E23, or 4.8E25 electronsMultiply this by 1E-12 to get 4.8E13 added electronsMultiply by 1.6E-19 to get 7.68E-6 coulombs (for the next part)2The mass of the robot head is irrelevantSince the charge is positive and the net sphere charge is negative, the force is attractiveThe force is the product of the charges (4.32E-11 coulombs^2) over the square of the distance (0.25), divided by (4 pi times the electric constant 8.85E-12), or 1.56 newtons 4The contact will discharge the positive charge, leaving a total net charge of 7.68E-6 - 5.63E-6 2.05E-6 coulombs, and replace it with some amount of negative chargeThe exact amount will depend on sizes and shapes (data not given), but for the sake of the problem we will suppose that it divides evenlyThat will put 1.02E-6 coulombs on each, and at a distance of 1 meter, the repulsive force will be 1.04E-12/(4 pi x 8.85E-12), or 9E-3 newtons.
