How many electrons must be removed from an electrically neutral dollar to give it a charge of 3.8 micro- Coulomb?

To find the number of electrons that must be removed from a dollar to give it a charge of 3.8 micro-Coulombs, we can use the formula:

Q = n * e

Where:
Q = charge (3.8 micro-Coulombs)
n = number of electrons
e = elementary charge (1.6 x 10^-19 Coulombs)

Plugging in the values:

3.8 x 10^-6 = n * 1.6 x 10^-19

n = (3.8 x 10^-6) / (1.6 x 10^-19)
n = 2.375 x 10^13

Therefore, approximately 2.375 x 10^13 electrons must be removed from a dollar to give it a charge of 3.8 micro-Coulombs.

To calculate the number of electrons that must be removed, we can use the following formula:

```
Number of electrons = Charge / (Charge of an electron)
```

The charge of an electron is:

```
-1.602 x 10^-19 Coulombs
```

So, the number of electrons that must be removed is:

```
Number of electrons = 3.8 x 10^-6 Coulombs / (-1.602 x 10^-19 Coulombs)
```

```
Number of electrons = -2371 electrons
```

Therefore, 2371 electrons must be removed from an electrically neutral dollar to give it a charge of 3.8 microCoulombs.