Calculating Current given the number of electrons and positive ions

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Homework Statement

In a fluorescent tube of diameter 3 cm, 10 *10¹⁸ electrons and 2.5 * 10¹⁸ positive ions (with a charge of +e) flow through a cross-sectional area each second. What is the current in the tube?

Homework Equations

Current is I = (delta Q)/(deta T)

The Attempt at a Solution

delta t is given as 1 second, but how do u calculate the change in charge given this information?

 

[Delphi51]

Use the charge on the electron. Note that the positive ions will go the opposite way to the electrons, so you must add the two currents.

 

[nlightner]

I would approach this problem by first understanding the basic principles of current and charge. Current is defined as the rate of flow of electric charge, and is measured in units of amperes (A). Charge, on the other hand, is measured in units of coulombs (C).

In this scenario, we are given the number of electrons and positive ions flowing through the cross-sectional area of the fluorescent tube per second. This tells us the total amount of charge passing through the area per second. To calculate the current, we need to determine the change in charge over time.

To do this, we can use the equation for current: I = (delta Q)/(delta t). In this case, delta t is given as 1 second, so we just need to calculate the change in charge, delta Q.

Since we are given the number of electrons and positive ions, we can use the fact that each electron has a charge of -e and each positive ion has a charge of +e. This means that the total charge passing through the area per second is (10 * 10^18 * -e) + (2.5 * 10^18 * +e) = -7.5 * 10^18 e.

To convert this to coulombs, we can use the fact that 1 e = 1.602 * 10^-19 C. This gives us a total charge of -7.5 * 10^18 * 1.602 * 10^-19 C = -12.03 C.

Now, we can plug this value for delta Q into the equation for current: I = (-12.03 C)/(1 s) = -12.03 A.

We must also consider the direction of the current. Since the charge carriers in this scenario are both positive and negative, the overall current will be the sum of the individual currents, taking into account their direction. In this case, the negative charge carriers (electrons) will contribute a negative current, while the positive charge carriers (positive ions) will contribute a positive current. So, the overall current in the tube will be -12.03 A + 12.03 A = 0 A.

In conclusion, the current in the fluorescent tube is 0 A, given the number of electrons and positive ions flowing through the cross-sectional area per second. This result is not surprising, as the total charge passing through the area