In a fluorescent tube of diameter 4.9 cm , 2.7 × 1018 electrons (with a charge o

Question

In a fluorescent tube of diameter 4.9 cm , 2.7 × 1018 electrons (with a charge of ?e) and 1.2 × 1018 positive ions (with a charge of +e) flow in opposite directions through a cross-sectional each second. What is the current in the tube? The fundamental charge is 1.602 × 10?19 C. Answer in units of A

009 5.0 points In a fluorescent tube of diameter 4.9 cm, 2.7 x 1018 electrons (with a charge of -e) and 1.2 x 10 positive ions (with a charge of te) flow in opposite directions through a cross-sectional each second 18 What is the current in the tube? The fundamental charge is 1.602 × 10-19 C. Answer in units of A

Explanation / Answer

Suppose that one coulomb of negative charge alone flow through a conductor and no positive charges flow.

That means a current of one A flow in the opposite direction.

This is equivalent to one coulomb of positive charge flowing through and there is no negative charge,

Now in addition to that one coulomb of positive charge flows. This is flowing in the current direction of the previous one.

Now the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.

They cannot cancel each other, because even before the current flow the conductor was neutral.

And in any neutral conductor the positive charge will always be equal to the negative charges.

The current in the given problem is

[1.2 + 2.7] x 1018 * 1.602 x 10-19 C/s

= 0.624 A

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