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Exercicios do Tipler Resolvidos (Volume 2, Capitulo 21 ao 41)

The charge distributions are shown in the diagram. Thus, the net force acting on a test charge at the midpoint of the. Only the lines shown in d satisfy this requirement.

Hence, the force on either sphere will increaseif a third uncharged metal ball is placed between misca. 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.

Parte 1 de 11 Chapter 21 The Electric Field 1: The sphere will be negatively charged. Because the field is nonuniform and is larger in the x direction, the force acting on the positive charge of the dipole in the direction of increasing x will be greater than the force acting on the negative charge of the dipole in the direction of decreasing x and thus there will be a net electric force on the dipole in the direction of increasing x.


Then use that negatively charged sphere to charge the second metal sphere positively by induction.

The gravitational constant G is many orders of magnitude smaller than the Coulomb constant k. Like charges repel; like masses attract. A positively charged ball will induce a dipole on the metal ball, and if the two are in close proximity, the net force can be attractive.

When S is opened, these charges are trapped on B and remain there fisiac the charged body is removed. There are positive and negative charges but only positive fjsica.

Hence B is negatively charged and correct. The reduction of an electric field by the alignment of dipole moments with the field is discussed in further kosca in Chapter If the metal balls are placed in water, the water molecules around each ball tend to align themselves with the electric field.

This is shown for the ball tipper the right with charge —q. Imagine a negative charge situated to its right and a larger positive charge on the same line and the right of the negative charge.

Tipler Vol 2. 6 Ed Cap 25

The force is directly proportional to the product of the charges or masses. E due to the charge —q on the ball on the right plus the field due to the layer of positive charge that surrounds the ball on the right. This is shown in the diagram. In order to charge a body by induction, it must have charges that are free to move about on the body. Determine the Concept Er is zero wherever the net force acting on a test charge is zero.


Tipler Vol 2. 6 Ed Cap 25

On the other sphere, the net charge is positive and on the side far from the rod. When the charged wand is brought near the tinfoil, the side nearer the wand becomes positively charged by induction, and so it swings toward the wand.

This electric field is directed to the right. An insulator does not have such charges. First charge one metal sphere negatively by induction as in a. Hence, the force will decrease when the balls are placed in the water. Tags Exercicios do Tipler Resolvidos. When it touches the wand, fisca of the negative charge is transferred to the foil, which, as a result, acquires a net negative charge and is now repelled by the wand.

In this situation, the net electric field at the location of the sphere on the left is due only to the charge —q on the sphere on the right. Such an arrangement ffisica charges, with the distances properly chosen, would result in a net force of zero acting on Q. The positive charge and the induced charge on the neutral conductor, being of opposite sign, will always attract one another.