Question

"A total charge of 9.0 mC passes through a wire in 3.5 seconds. What is the current in the wire? How many electrons pass through the cross-sectional area in 10.0 s? If the number of charges that pass through the cross-sectional area during the given time interval doubles, what is the resulting current?"

Answer 1

1) 0.00257 A

2) $1.6\cdot 10^{17}$ electrons

3) 0.00512 A

Explanation:

1)

A current is defined as the flow of charge through a conductor.

The intensity of current is calculated as:

$I=\frac{q}{t}$ (1)

where

q is the amount of charge passing through a certain point in the conductor

t is the time interval during which this charge passes

In the wire in this problem we have:

$q=9.0 mC=0.009 C$ is the charge

$t=3.5 s$ is the time elapsed

Therefore, the current in the wire is:

$I=\frac{0.009}{3.5}=0.00257 A$

2)

To find the total charge passing through a given point of the wire in a certain time, we re-arrange eq. (1):

$q=It$

where

I is the current

t is the time interval we are considering

In this problem we have:

I = 0.00257 A is the current in the wire

t = 10.0 s is the time we are considering

Therefore, the charge is:

$q=(0.00257)(10.0)=0.0257 C$

We know that this charge q consists of N electrons, so we can write

$q=Ne$

where

$e=1.6\cdot 10^{-19}C$ is the charge of one electron

Solving for N, we find:

$N=\frac{q}{e}=\frac{0.0257}{1.6\cdot 10^{-19}}=1.6\cdot 10^{17}$

3)

In this problem, we are told that the number of charges that pass through the cross-sectional area during the given time interval doubles, so we have:

$N'=2N=2(1.6\cdot 10^{17})=3.2\cdot 10^{17}$

Therefore, the total charge through the point in the wire in a time of

t = 10.0 s

will be

$q'=N'e=(3.2\cdot 10^{17})(1.6\cdot 10^{-19})=0.0512 C$

And so, the current in this case will be

$I'=\frac{q'}{t}=\frac{0.0512}{10.0}=0.00512 A$

And we see that this current is twice the current we had in part 1), because the current is proportional to the number of charge carriers.

Answer 2

The charges are  $0.00257\ A$,  $1.6 \times 10^{17}$ electrons, and $0.00512 \ A$, and their further calculation can be defined as follows:

Current calculation:

For part 1:

A movement of energy via a conductor is known as just a current.

This current's strength is computed as:

$I=\frac{q}{t}.....(1)$

wherein q is the quantity of charge travelling through a specific location in the conductor, and t denotes a time interval during which this charge flows.

In the this problem, we have the following wire:

$\to q=9.0 \ m C=0.009\ C$ charges  

Time elapsed $t=3.5 \ s$

current wire:

$\to I=\frac{0.009}{3.5}= 0.00257 \ A$

For part 2:

You re-arrange eq. (1) to obtain that total charge passing through a particular location of the wire inside a given time.

$\to q=It$

where

current =I

interval time =t

$I = 0.00257\ A$  current in the wire

$t = 10.0\ s$

Calculating the charge:

$\to q=(0.00257) (10.0) =0.02575\ C$

We assume there are N electrons in this test charge, thus we can write

$q=Ne$

where

Calculating the charge in one electron

$e=1.6 \times 10^{-19}\ C$

Solving for N:

$\to N=\frac{q}{e}=\frac{0.0257}{1.6\times 10^{-19}}=1.6 \times 10^{17}$

For point 3:

A number of charges which it pass through the cross-sectional area during the given time interval doubles in this issue, so we have:

$\to N'=2N=2(1.6 \times 10^{17})=3.2\times 10^{17}$

As a result, the complete charge passed through the wire point in a period of

$t = 10.0\ s$

$\to q'=N'e=(3.2\times 10^{17})(1.6\times 10^{-19})=0.0512\ C$

So, in this scenario, the current will be

$\to I'=\frac{q'}{t}=\frac{0.0512}{10.0}=0.00512\ A$

Because the current is proportional to the number of charge carriers, we can see that this current is twice as large as the current we had in part 1.

Find out more about the total charge here:

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