HTTP://PHYSICSACT.WORDPRESS.COMChapter 21 The Electric Field 1: Discrete Charge DistributionsConceptual Problems.1. Similarities: The force between charges and masses varies as 1/r2.
The force is directly proportional to the product of the charges or masses.Differences: There are positive and negative charges but only positive masses. Like charges repel; like masses attract.Thegravitational constant G is many orders of magnitude smaller than the Coulomb constant k. 2. Determine the Concept No. In order to charge a body by induction, it must have charges that are free to move about on the body. An insulator does not have such charges.
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3. Determine the Concept During this sequence of events, negative charges are attracted from ground to the rectangular metal plate B.When S is opened, these charges are trapped on B and remain there when the charged body is removed. Hence B is negatively charged and (c) is correct. 4.
(a) Connect the metal sphere to ground; bring the insulating rod near the metal sphere and disconnect the sphere from ground; then remove the insulating rod. The sphere will be negatively charged. (b) Bring the insulating rod in contact with themetal sphere; some of the positive charge on the rod will be transferred to the metal sphere.
First charge one metal sphere negatively by induction as in (a). Then use that negatively charged sphere to charge the second metal sphere positively by induction.12Chapter 21.5. Determine the Concept Because the spheres are conductors, there are free electrons on them thatwill reposition themselves when the positively charged rod is brought nearby.
(a) On the sphere near the positively charged rod, the induced charge is negative and near the rod. On the other sphere, the net charge is positive and on the side far from the rod.
This is shown in the diagram. (b) When the spheres are separated and far apart and the rod has been removed, the induced charges aredistributed uniformly over each sphere. The charge distributions are shown in the diagram. 6. Determine the Concept The forces acting on +q are shown in the diagram. The force acting on +q due to −Q is along the line joining them and directed toward −Q. The force acting on +q due to +Q is along the line joining them and directed away from +Q.
Because charges +Q and −Q are equal in magnitude, the forcesdue to these charges are equal and their sum (the net force on +q) will be to the right and so (e) is correct. Note that the vertical components of these forces add up to zero.7. Determine the Concept The acceleration of the positive charge is given byr r F q0 r a= = E. Because q0 and m are both positive, the acceleration is in the same m m direction as the electric field. (d ) is correct.8.
r Determine the Concept E is zero wherever the net force acting on a test charge is zero. At the center of the square the two positive charges alone would produce a net electric field of zero, and the two negative charges alone would also produce a net electric field of zero. Thus, the net force acting on a test charge at the midpoint of theThe Electric Field 1: Discrete ChargeDistributionssquare will be zero. (b) is correct. 9. (a) The zero net force acting on Q could be the consequence of equal collinear charges being equidistant from and on opposite sides of Q.
(b) The charges described in (a) could be either positive or negative and the net force on Q would still be zero.3(c) Suppose Q is positive. Imagine a negative charge situated to its right and a largerpositive charge on the same line and the right of the negative charge. Such an arrangement of charges, with the distances properly chosen, would result in a net force of zero acting on Q. (d) Because none of the above are correct, (d ) is correct. 10. Determine the Concept We can use the rules for drawing electric field lines to draw the electric field lines for this system.
In the sketch to the.