Suppose that `1.0g` fo hydrogen is separated into electrons and protons. Suppose also that the protons are palced at the earth's north pole and the electrons are placed at the south pole. What is the resulting compression force on earth? (Given that the radius of rarth is 6400 km).

Explanation

To solve the problem of finding the resulting compression force on Earth when 1.0 g of hydrogen is separated into electrons and protons, we will follow these steps:

### Step 1: Determine the number of protons and electrons in 1.0 g of hydrogen.
Hydrogen (H) has a molar mass of approximately 1 g/mol. Since we have 1.0 g of hydrogen, we can find the number of moles:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.0 \, \text{g}}{1 \, \text{g/mol}} = 1 \, \text{mol} \]

Since each molecule of H₂ contains 2 hydrogen atoms, we have:

\[ \text{Number of hydrogen atoms} = 2 \times 1 \, \text{mol} = 2 \, \text{mol} \]

Using Avogadro's number (\(6.022 \times 10^{23} \, \text{atoms/mol}\)), we can find the total number of protons and electrons:

\[ \text{Number of protons} = 2 \times 6.022 \times 10^{23} \approx 1.2044 \times 10^{24} \]
\[ \text{Number of electrons} = 2 \times 6.022 \times 10^{23} \approx 1.2044 \times 10^{24} \]

### Step 2: Calculate the total charge of protons and electrons.
The charge of one proton is approximately \(+1.6 \times 10^{-19} \, \text{C}\) and the charge of one electron is approximately \(-1.6 \times 10^{-19} \, \text{C}\).

Total charge of protons (\(Q_p\)):
\[ Q_p = \text{Number of protons} \times \text{Charge of one proton} = 1.2044 \times 10^{24} \times 1.6 \times 10^{-19} \approx 1.927 \times 10^{5} \, \text{C} \]

Total charge of electrons (\(Q_e\)):
\[ Q_e = \text{Number of electrons} \times \text{Charge of one electron} = 1.2044 \times 10^{24} \times (-1.6 \times 10^{-19}) \approx -1.927 \times 10^{5} \, \text{C} \]

### Step 3: Calculate the distance between the charges.
The distance between the north pole and south pole of the Earth is approximately twice the radius of the Earth. Given that the radius of the Earth is \(6400 \, \text{km}\):

\[ \text{Distance} = 2 \times 6400 \, \text{km} = 12800 \, \text{km} = 12800 \times 10^3 \, \text{m} = 1.28 \times 10^7 \, \text{m} \]

### Step 4: Use Coulomb's Law to calculate the force.
Coulomb's Law states that the force \(F\) between two charges is given by:

\[ F = k \frac{|Q_p \cdot Q_e|}{r^2} \]

Where \(k\) (Coulomb's constant) is approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).

Substituting the values:

\[ F = 8.99 \times 10^9 \frac{|1.927 \times 10^5 \cdot (-1.927 \times 10^5)|}{(1.28 \times 10^7)^2} \]

Calculating the denominator:

\[ (1.28 \times 10^7)^2 = 1.6384 \times 10^{14} \]

Now substituting back into the force equation:

\[ F = 8.99 \times 10^9 \frac{(1.927 \times 10^5)^2}{1.6384 \times 10^{14}} \]

Calculating \((1.927 \times 10^5)^2\):

\[ (1.927 \times 10^5)^2 \approx 3.709 \times 10^{10} \]

Now substituting this value:

\[ F = 8.99 \times 10^9 \frac{3.709 \times 10^{10}}{1.6384 \times 10^{14}} \approx 2.06 \times 10^5 \, \text{N} \]

### Final Answer:
The resulting compression force on Earth is approximately \(2.06 \times 10^5 \, \text{N}\).

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