Two copper balls, each weighing `10 g` are kept in air `10 cm` apart. If one electron from every `10^(6)` atoms in trandferred from one ball to the other, the coulomb force between them is (atomic weight of copper is `63.5`)

Options

A. `2.0 xx 10^(10)N`
B. `2.0 xx 10^(4) N`
C. `2.0 xx 10^(8)N`
D. `2.0 xx 10^(6)N`

Answer

To solve the problem of finding the Coulomb force between two copper balls after transferring electrons, we can follow these steps:

### Step 1: Calculate the number of atoms in one copper ball
Given:
- Mass of one copper ball (m) = 10 g
- Atomic weight of copper (A) = 63.5 g/mol
- Avogadro's number (N_A) = \(6.02 \times 10^{23}\) atoms/mol

The number of moles of copper in one ball can be calculated using:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{atomic weight}} = \frac{10 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.1575 \, \text{mol} \]

Now, we can find the number of atoms:
\[ \text{Number of atoms} = \text{Number of moles} \times N_A = 0.1575 \times 6.02 \times 10^{23} \approx 9.48 \times 10^{22} \, \text{atoms} \]

### Step 2: Determine the number of electrons transferred
Given that one electron is transferred from every \(10^6\) atoms, the total number of electrons transferred (n) can be calculated as:
\[ n = \frac{\text{Number of atoms}}{10^6} = \frac{9.48 \times 10^{22}}{10^6} \approx 9.48 \times 10^{16} \, \text{electrons} \]

### Step 3: Calculate the charge on each ball
The charge of one electron (e) is:
\[ e = 1.6 \times 10^{-19} \, \text{C} \]

The total charge (Q) gained by each ball is:
\[ Q = n \times e = 9.48 \times 10^{16} \times 1.6 \times 10^{-19} \approx 0.015 \, \text{C} \]

### Step 4: Calculate the Coulomb force between the two balls
The formula for the Coulomb force (F) between two charges is given by:
\[ F = k \frac{Q_1 Q_2}{r^2} \]
where:
- \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) (Coulomb's constant)
- \(Q_1 = Q_2 = Q = 0.015 \, \text{C}\)
- \(r = 10 \, \text{cm} = 0.1 \, \text{m}\)

Substituting the values:
\[ F = 9 \times 10^9 \frac{(0.015)^2}{(0.1)^2} \]
\[ F = 9 \times 10^9 \frac{0.000225}{0.01} = 9 \times 10^9 \times 0.0225 = 2.025 \times 10^8 \, \text{N} \]

Thus, the Coulomb force between the two balls is approximately:
\[ F \approx 2.0 \times 10^8 \, \text{N} \]

### Final Answer:
The Coulomb force between the two copper balls is \(2.0 \times 10^8 \, \text{N}\).

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