A particle of mass `1.96xx10^(-15)` kg is kept in equilibrium between two horizontal metal plates having potential difference of 400 V separated apart by 0.02 m. then the charge on the particle is (e= electronic charge)

Options

A. `3 e`
B. `6 e`
C. `2 e`
D. `5 e`

Answer

To find the charge on the particle, we can follow these steps:

### Step 1: Calculate the Electric Field (E)
The electric field (E) between two parallel plates can be calculated using the formula:
\[ E = \frac{\Delta V}{d} \]
where \(\Delta V\) is the potential difference and \(d\) is the separation between the plates.

Given:
- \(\Delta V = 400 \, V\)
- \(d = 0.02 \, m\)

Substituting the values:
\[ E = \frac{400 \, V}{0.02 \, m} = 20000 \, \text{N/C} = 2 \times 10^4 \, \text{N/C} \]

### Step 2: Set Up the Force Equilibrium
Since the particle is in equilibrium, the electric force (F_E) acting on it due to the electric field must balance the gravitational force (F_g) acting on it.

The gravitational force is given by:
\[ F_g = mg \]
where:
- \(m = 1.96 \times 10^{-15} \, kg\)
- \(g = 9.8 \, m/s^2\)

Calculating \(F_g\):
\[ F_g = 1.96 \times 10^{-15} \, kg \times 9.8 \, m/s^2 = 1.92 \times 10^{-14} \, N \]

### Step 3: Relate Electric Force to Charge
The electric force (F_E) acting on the particle can be expressed as:
\[ F_E = QE \]
where \(Q\) is the charge on the particle.

Since \(F_E = F_g\), we can write:
\[ QE = mg \]

### Step 4: Solve for Charge (Q)
Now, substituting the values we have:
\[ Q \cdot (2 \times 10^4 \, \text{N/C}) = 1.92 \times 10^{-14} \, N \]
Solving for \(Q\):
\[ Q = \frac{1.92 \times 10^{-14} \, N}{2 \times 10^4 \, \text{N/C}} = 9.6 \times 10^{-19} \, C \]

### Step 5: Express Charge in Terms of Electronic Charge (e)
The charge of an electron \(e\) is approximately:
\[ e = 1.6 \times 10^{-19} \, C \]
Now, we can express \(Q\) in terms of \(e\):
\[ Q = \frac{9.6 \times 10^{-19} \, C}{1.6 \times 10^{-19} \, C} = 6 \, e \]

### Conclusion
Thus, the charge on the particle is \(6\) times the electronic charge.

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