How many electrons should be removed from a coin of mass `1.6g`, so that it may float in an electric field of intensity of `10^(9)NC^(-1)` directed upward. (take `g= 10m//s^(2)`)

Options

A. `10^(6)`
B. `10^(7)`
C. `10^(9)`
D. `10^(8)`

Explanation

To determine how many electrons should be removed from a coin of mass 1.6 g so that it may float in an electric field of intensity \(10^9 \, \text{N/C}\), we can follow these steps:

### Step 1: Calculate the weight of the coin
The weight \(W\) of the coin can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \(m = 1.6 \, \text{g} = 0.0016 \, \text{kg}\) (converting grams to kilograms)
- \(g = 10 \, \text{m/s}^2\)

Substituting the values:
\[ W = 0.0016 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 0.016 \, \text{N} \]

### Step 2: Calculate the electric force needed to balance the weight
For the coin to float in the electric field, the electric force \(F_e\) acting on it must equal its weight \(W\):
\[ F_e = W = 0.016 \, \text{N} \]

### Step 3: Relate electric force to charge and electric field
The electric force can also be expressed as:
\[ F_e = q \cdot E \]
where:
- \(q\) is the charge on the coin,
- \(E = 10^9 \, \text{N/C}\) is the electric field intensity.

Setting the two expressions for electric force equal gives:
\[ q \cdot E = 0.016 \, \text{N} \]

### Step 4: Solve for the charge \(q\)
Rearranging the equation to solve for \(q\):
\[ q = \frac{0.016 \, \text{N}}{10^9 \, \text{N/C}} = 1.6 \times 10^{-11} \, \text{C} \]

### Step 5: Calculate the number of electrons to be removed
The charge of a single electron \(e\) is approximately \(1.6 \times 10^{-19} \, \text{C}\). To find the number of electrons \(n\) that corresponds to the charge \(q\), we use:
\[ n = \frac{q}{e} = \frac{1.6 \times 10^{-11} \, \text{C}}{1.6 \times 10^{-19} \, \text{C}} = 10^8 \]

### Final Answer
Thus, the number of electrons that should be removed from the coin is:
\[ \boxed{10^8} \]