Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

In the Faraday pail experiment a metal ball is lowered into a brass pail. In a variation of this experiment, suppose we charge the metal ball with Q and the brass pail with charge 2Q. The ball is slowly lowered into the pail. At no time does the sphere touch the pail. While the ball is inside, the charge on the outside of the pail is _____. The outer part of the pail is then touched/grounded. Then the ball is removed. After the ball is removed, the charge on the outer surface of the pail is ___. Briefly support your answers with reasoning from Gauss's Law.

Sagot :

Answer:

a)

*The charges of which and the cube are of the same sign..

          Q_exterior = 3 Q

* the charge of the sphere has a different sign than the charge of the cube,

         Q_exterior =  Q

b)   Q = 0

Explanation:

To correctly describe this experiment, you must remember that in metals charges are mobile and that charges of the same sign repel and of different signs attract.

Let's analyze each situation

a) Suppose that the charges of which and the cube are of the same sign.

When the ball is introduced without touching the walls, its charge Q attracts a charge of equal magnitude and different sign to the internal wall. If we create a Gaussian surface around the inner wall of the sphere the net charge between the ball and the inner wall is zero. Consequently, a face Q should have been generated in the outer wall, therefore in this wall it has a total load of

               Q_exterior = 3 Q

   Now suppose that the charge of the sphere has a different sign than the charge of the cube, for simplicity let's say that the charge of the sphere is -Q and the cube + 2Q,

     Again we create a Gaussian surface outside the inner wall, now the charge on the ball attracts a charge of value + Q to neutralize the charge between the ball and the inner wall. Therefore a load remains on the outer wall

              Q_exterior = + Q

b) The cube is connected to earth and it is touched with the ball, in this case the charge of the two bodies is neutralized by the Earth, therefore the bodies have zero charge

             Q = 0

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.