An `alpha` particle is situated in an electric field of `10^(6)N//C`. The force exerted or it is

Options

A. `6.4 xx 10^(-3)N`
B. `3.2 xx 10^(-13)N`
C. `32 xx 10^(-13)N`
D. `64 xx 10^(-13)N`

Answer

To solve the problem of finding the force exerted on an alpha particle situated in an electric field of \(10^6 \, \text{N/C}\), we can follow these steps:

### Step 1: Understand the charge of the alpha particle
An alpha particle consists of 2 protons and 2 neutrons. Since neutrons do not carry a charge, we only consider the protons. The charge of an alpha particle is equal to the charge of 2 protons.

- The charge of one proton is approximately \(1.6 \times 10^{-19} \, \text{C}\).
- Therefore, the charge of an alpha particle \(q\) is:
\[ q = 2 \times (1.6 \times 10^{-19}) = 3.2 \times 10^{-19} \, \text{C} \]

### Step 2: Use the formula for electric force
The force \(F\) exerted on a charge in an electric field is given by the formula:
\[ F = q \times E \]
where:
- \(F\) is the force,
- \(q\) is the charge,
- \(E\) is the electric field strength.

### Step 3: Substitute the values into the formula
We know:
- \(q = 3.2 \times 10^{-19} \, \text{C}\)
- \(E = 10^6 \, \text{N/C}\)

Now we can substitute these values into the formula:
\[ F = (3.2 \times 10^{-19}) \times (10^6) \]

### Step 4: Calculate the force
Now, we perform the multiplication:
\[ F = 3.2 \times 10^{-19} \times 10^6 = 3.2 \times 10^{-13} \, \text{N} \]

### Final Answer
The force exerted on the alpha particle is:
\[ F = 3.2 \times 10^{-13} \, \text{N} \]

---