An ebonite rod acquires a negative charge of `3.2 xx 10^(-10)` C. The number of excess electrons it has is given by

Options

A. `2 xx 10^9`
B. `2 xx 10^(-9)`
C. `2 xx 10^(-29)`
D. `2 xx 10^(29)`

Explanation

To find the number of excess electrons on an ebonite rod that has acquired a negative charge of \(3.2 \times 10^{-10}\) C, we can follow these steps:

### Step 1: Understand the relationship between charge and the number of electrons
The total charge \(Q\) acquired by the rod can be expressed in terms of the number of excess electrons \(n\) and the charge of a single electron \(e\):
\[ Q = n \cdot e \]
where:
- \(Q\) is the total charge,
- \(n\) is the number of excess electrons,
- \(e\) is the charge of a single electron, which is approximately \(1.6 \times 10^{-19}\) C.

### Step 2: Rearrange the formula to find \(n\)
We can rearrange the formula to solve for \(n\):
\[ n = \frac{Q}{e} \]

### Step 3: Substitute the known values
Now, we can substitute the known values into the equation:
- \(Q = -3.2 \times 10^{-10}\) C (the negative sign indicates excess electrons),
- \(e = 1.6 \times 10^{-19}\) C.

Thus, we have:
\[ n = \frac{-3.2 \times 10^{-10}}{-1.6 \times 10^{-19}} \]

### Step 4: Perform the calculation
Now we can calculate \(n\):
\[ n = \frac{3.2 \times 10^{-10}}{1.6 \times 10^{-19}} = \frac{3.2}{1.6} \times \frac{10^{-10}}{10^{-19}} \]
Calculating the fraction:
\[ \frac{3.2}{1.6} = 2 \]
And for the powers of ten:
\[ \frac{10^{-10}}{10^{-19}} = 10^{9} \]
So, we have:
\[ n = 2 \times 10^{9} \]

### Step 5: Conclusion
The number of excess electrons on the ebonite rod is:
\[ n = 2 \times 10^{9} \]