6.	An amoeba has 1.00×1016 protons and a net charge of 0.300 pC.
a.	 How many fewer electrons are there than protons?
b.	 If you paired them up, what fraction of the protons would have no electrons?

### Part a: How many fewer electrons are there than protons?

The number of protons in the amoeba is given as \( N_p = 1.00 \times 10^{16} \). The net charge of the amoeba is \( 0.300 \, \text{pC} = 0.300 \times 10^{-12} \, \text{C} \).

Since protons have a positive charge and electrons have a negative charge, the net charge of the amoeba is due to the difference between the number of protons and electrons.

The charge of a single proton or electron is \( e = 1.602 \times 10^{-19} \, \text{C} \).

Let \( N_e \) be the number of electrons in the amoeba. The net charge \( Q \) is given by the difference in charge between protons and electrons:
\[
Q = (N_p - N_e) \times e
\]
Substitute the known values into the equation:
\[
0.300 \times 10^{-12} \, \text{C} = (1.00 \times 10^{16} - N_e) \times 1.602 \times 10^{-19} \, \text{C}
\]
Solve for \( N_e \).

### Part b: What fraction of the protons would have no electrons?

Let \( \Delta N = N_p - N_e \) be the difference between the number of protons and electrons, which is the number of protons without paired electrons. The fraction of protons without electrons is:
\[
\text{Fraction} = \frac{\Delta N}{N_p}
\]
Using the values from part a, this can be calculated.

Let's first solve for \( N_e \) and \( \Delta N \).# Given values
Q = 0.300e-12  # Net charge in Coulombs (0.300 pC)
Np = 1.00e16  # Number of protons
e = 1.602e-19  # Charge of a single proton or electron in Coulombs

# Using the formula: Q = (Np - Ne) * e, solve for Ne
Ne = Np - (Q / e)  # Number of electrons

# Difference in number of protons and electrons
delta_N = Np - Ne  # Number of fewer electrons

# Fraction of protons that have no electrons
fraction_protons_no_electrons = delta_N / Np

Ne, delta_N, fraction_protons_no_electrons### Part a: How many fewer electrons are there than protons?

There are approximately \( 1,872,660 \) fewer electrons than protons in the amoeba.

### Part b: What fraction of the protons would have no electrons?

The fraction of protons without paired electrons is approximately \( 1.87 \times 10^{-10} \). This means a very small fraction of the protons (0.0000000187%) are unpaired.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. How does the concept of charge neutrality apply to this problem?
2. What would happen if the amoeba gained more electrons?
3. How is the electric field affected by the excess protons?
4. Can the net charge affect the chemical properties of the amoeba?
5. How would this calculation change if we dealt with different ions?

**Tip**: When solving physics problems, always keep track of the units; they help ensure accuracy and clarity.

Explore how to calculate the number of electrons in an amoeba based on its net charge and proton count. Understand the concepts of charge quantization and how to determine the fraction of protons without paired electrons. Suitable for advanced high school students studying electrostatics and charge interactions.

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