The electric field that can balance a charged particle of mass `3.2xx10^(-27)`kg is (Given that the charge on the particle is `1.6xx10^(-19)C`)

Options

A. `19.6xx10^(-8)NC^(-1)`
B. `20xx10^(-6)NC^(-1)`
C. `19.6xx10^(8)NC^(-1)`
D. `20xx10^(6)NC^(-1)`

Explanation

To find the electric field that can balance a charged particle of mass \(3.2 \times 10^{-27}\) kg with a charge of \(1.6 \times 10^{-19}\) C, we can follow these steps:

### Step 1: Understand the forces acting on the charged particle
The charged particle experiences two forces:
1. The gravitational force (\(F_g\)) acting downwards, given by:
\[ F_g = mg \]
where \(m\) is the mass of the particle and \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).

2. The electric force (\(F_e\)) acting upwards, given by:
\[ F_e = qE \]
where \(q\) is the charge of the particle and \(E\) is the electric field.

### Step 2: Set the forces equal for balance
For the particle to be in equilibrium (balanced), the upward electric force must equal the downward gravitational force:
\[ F_e = F_g \]
This can be expressed as:
\[ qE = mg \]

### Step 3: Rearrange the equation to solve for the electric field \(E\)
We can rearrange the equation to solve for \(E\):
\[ E = \frac{mg}{q} \]

### Step 4: Substitute the known values
Now, we will substitute the known values into the equation:
- Mass \(m = 3.2 \times 10^{-27} \, \text{kg}\)
- Charge \(q = 1.6 \times 10^{-19} \, \text{C}\)
- Acceleration due to gravity \(g = 9.8 \, \text{m/s}^2\)

Substituting these values:
\[ E = \frac{(3.2 \times 10^{-27} \, \text{kg})(9.8 \, \text{m/s}^2)}{1.6 \times 10^{-19} \, \text{C}} \]

### Step 5: Calculate the numerator
Calculating the numerator:
\[ 3.2 \times 10^{-27} \times 9.8 = 31.36 \times 10^{-27} \, \text{kg m/s}^2 \]

### Step 6: Calculate the electric field \(E\)
Now, substituting back into the equation for \(E\):
\[ E = \frac{31.36 \times 10^{-27}}{1.6 \times 10^{-19}} \]

### Step 7: Simplify the fraction
Calculating the fraction:
\[ E = 19.6 \times 10^{-8} \, \text{N/C} \]

### Final Answer
Thus, the electric field required to balance the charged particle is:
\[ E = 19.6 \times 10^{-8} \, \text{N/C} \]

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