Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 8.7 nC. Using insulating handles, the blocks are moved so they touch each other. After touching for a few seconds, the blocks are separated (again using insulating handles).(a) What is the final charge of block A?
(b) What happened while the blocks were in contact with each other?

Respuesta :

Explanation:

(a)   Since, it is given that the blocks are identical so distribution of charge will be uniform on both the blocks.

Hence, final charge on block A will be calculated as follows.

         Charge on block A = [tex]\frac{(8.7 + 0 nC}{2}[/tex]

                                           = 4.35 nC

Therefore, final charge on the block A is 4.35 nC.

(b)  As it is given that the positive charge is coming on block A . This means that movement of electrons will be from A to B.

Thus, we can conclude that while the blocks were in contact with each other then electrons will flow from A to B.

Answer:

a) final charge on block A, [tex]Q_A=4.35\times 10^{-9}\ C[/tex]

b) There is a flow of current when the two blocks were brought in contact since there is imbalance of the charge distribution and both are identical metal masses.

Explanation:

Given:

  • initial charge on metal block A, [tex]Q_a=0\ C[/tex]
  • initial charge on metal block B, [tex]Q_b=8.7\times 10^{-9}\ C[/tex]

After the two are brought in contact with each other final charge on each of the block will be equal because the two bodies are identical.

So,

a)

Final charge on block A and block B, [tex]Q_A=Q_B=\frac{8.7\times 10^{-9}}{2} =4.35\times 10^{-9}\ C[/tex]

b)

When the two blocks are brought into physical contact of together then the charges from the charged block move to the neutral block until there are equal charges on both the identical blocks. This flow of charge creates a flow of current for a short time till the charges get balanced on both the blocks to attain equilibrium. Charges flow due to the potential difference.