An ebonite rod acquires a negative change of `6.4 xx 10^-10 C`. The number of excess electrons it has is

Options

A. `4 xx 10^9`
B. `4 xx 10^(-9)`
C. `4 xx 10^(-29)`
D. `4 xx 10^(29)`

Explanation

To find the number of excess electrons on an ebonite rod that acquires a negative charge of \(6.4 \times 10^{-10} \, C\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Charge of an Electron:**
The charge of a single electron is approximately \(e = -1.6 \times 10^{-19} \, C\).

2. **Use the Formula for Charge:**
The total charge \(Q\) acquired by the ebonite rod can be expressed in terms of the number of excess electrons \(n\) as:
\[ Q = n \cdot e \]
Since the charge is negative, we can write:
\[ -6.4 \times 10^{-10} \, C = n \cdot (-1.6 \times 10^{-19} \, C) \]

3. **Solve for \(n\):**
Rearranging the equation to solve for \(n\):
\[ n = \frac{Q}{e} \]
Substituting the values:
\[ n = \frac{-6.4 \times 10^{-10} \, C}{-1.6 \times 10^{-19} \, C} \]

4. **Calculate \(n\):**
Performing the division:
\[ n = \frac{6.4 \times 10^{-10}}{1.6 \times 10^{-19}} \]
To simplify this, we can divide the coefficients and subtract the exponents:
\[ n = 4 \times 10^{9} \]

5. **Conclusion:**
The number of excess electrons on the ebonite rod is:
\[ n = 4 \times 10^{9} \]

### Final Answer:
The ebonite rod has \(4 \times 10^{9}\) excess electrons.

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