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Practice Problems with Solution Key about Coulomb’s Law.
Typology: Exercises
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Last chapter, we learned that masses attract each other. We learned that the attractive force between two masses could be found using Newton’s Universal Law of Gravitation (Newton’s ULOG), Fg
GmM R^2
, where R is the distance between the centers of the objects. “G” was simply a constant, and its units could easily be found by solving the above equation for G.
When dealing with charged objects, we also talk of forces between them. Unlike last chapter, these forces can be EITHER attractive (for unlike charges) OR repulsive (for like charges). To find the force between charged objects, we can use Coulomb’s Law, which is 2 _d_ _kqQ Fe_. In this equation, “d” is the distance between the objects, q and Q are the charges on the charged objects, and k is a constant equal to _k_ 8.99 109 _N m_^2 _C_^2 . This law is similar in form and structure to Newton’s ULOG, and the relationships that we used last chapter still apply. For example, _Fe q_ (ex: If you double the charge on an object, you double the force) _Fe Q_ (ex: If you quarter the charge on an object, you quarter the force) _Fe qQ_ (ex: If you double both charges, you quadruple the force) 2 _d_ _Fe_ (ex: If you double the distance between the charges, you quarter the force. If you divide the distance between the charges by 3, you multiply the force by 9 times).
NEVER, EVER plug signs into Coulomb’s law. If two charged objects have opposite signs, plug only their magnitudes into the equation. At the end, make sure to account for the fact that they have different signs by saying that the forces were either “repulsive”. If the charges had opposite signs, you account for this by saying that the charges
### It is important to see that we can easily use q ne (or Q ne ) and 2 _d_ _kqQ Fe_ together in problem solving. If you are told how many electrons an object gains or loses, you can easily calculate “q” or “Q” to plug into Coulomb’s Law. ~~~~ Similarly to Newton’s ULOG, Coulomb’s Law is a vector law. When multiple charges are involved, make sure to draw the forces acting on each charged object (as vector arrows). Again, do not put signs into Coulomb’s law, as the vector arrows provide sufficient direction. Once the object has all the forces draw on it (free-body diagram style), then the forces can be added as vectors. ## Coulomb’s Law K = 8.99E9 N-m^2 /C^2 e = 1.602E-19 C me = 9.11E-31 kg mp = 1.67E-27 kg 1. Two charged spheres 10 cm apart attract each other with a force of 3.0 x 10 6 N. What force results from each of the following changes, considered separately? a) Both charges are doubled and the distance remains the same. b) An uncharged, identical sphere is touched to one of the spheres, and then taken far away. c) The separation is increased to 30 cm. 2. The force of electrostatic repulsion between two small positively charged objects, A and B, is 3.6 x 10 5 N when AB = 0.12m. What is the force of repulsion if AB is increased to a) 0.24 m b) 0.36 m 3. Calculate the force between charges of 5.0 x 10 8 C and 1.0 x 10 7 C if they are 5.0 cm apart. 4. What is the magnitude of the force a 1.5 x 10 6 C charge exerts on a 3.2 x 10 4 C charge located 1.5 m away? 5. Two spheres; 4.0 cm apart, attract each other with a force of 1.2 x 10 9 N. Determine the magnitude of the charge on each, if one has twice the charge (of the opposite sign) as the other. 6. Two equal charges of magnitude 1.1 x 10 7 C experience an electrostatic force of 4.2 x 10 4 N. How far apart are the centers of the two charges? 7. How many electrons must be removed from a neutral, isolated conducting sphere to give it a positive charge of 8.0 x 10 8 C? 8. What will be the force of electric repulsion between two small spheres placed 1.0 m apart, if each has a deficit of 10^8 electrons? 9. Two identical, small spheres of mass 2.0 g are fastened to the ends of a 0.60m long light, flexible, insulating fishing line. The fishing line is suspended by a hook in the ceiling at its exact centre. The spheres are each given an identical electric charge. They are in static equilibrium, with an angle of 30 between the string halves, as shown. Calculate the magnitude of the charge on each sphere_. (Hint: start off by drawing a FULL, DETAILED_ _FBD of one of the charged spheres)._ .30 m Coulomb’s Law Problems