An oil drop having a mass `4.8xx10^(-10)`g. and charge `2.4xx10^(-18)`C stand still between two charged horizontal plates seperated by a distance of `1cm`. If now the polarity of the plates is changed the instantaneous acceleration of the drop is `("in" ms^(2)),(g=10ms^(-2))`

Options

A. `5`
B. `10`
C. `20`
D. `40`

Explanation

To solve the problem step by step, we will analyze the forces acting on the oil drop and calculate the instantaneous acceleration after the polarity of the plates is changed.

### Step 1: Identify the forces acting on the oil drop
The oil drop has two forces acting on it:
1. **Gravitational Force (mg)**: This force acts downwards and is given by the formula:
\[ F_g = m \cdot g \]
2. **Electric Force (F_e)**: This force acts upwards when the oil drop is stationary between the plates. It is given by:
\[ F_e = E \cdot q \]
where \(E\) is the electric field strength and \(q\) is the charge of the oil drop.

### Step 2: Set up the equilibrium condition
Since the oil drop is initially at rest, the forces are balanced:
\[ F_e = F_g \]
Thus, we can write:
\[ E \cdot q = m \cdot g \]

### Step 3: Change the polarity of the plates
When the polarity of the plates is changed, the direction of the electric force reverses. Therefore, the electric force now acts downwards, in the same direction as the gravitational force.

### Step 4: Calculate the total force acting on the oil drop
Now, the total force acting on the oil drop when the polarity is changed becomes:
\[ F_{\text{total}} = F_e + F_g \]
Substituting the expressions for the forces:
\[ F_{\text{total}} = E \cdot q + m \cdot g \]
Since \(E \cdot q = m \cdot g\), we can substitute:
\[ F_{\text{total}} = mg + mg = 2mg \]

### Step 5: Relate the total force to acceleration
According to Newton's second law:
\[ F_{\text{total}} = m \cdot a \]
Substituting for \(F_{\text{total}}\):
\[ 2mg = m \cdot a \]

### Step 6: Solve for acceleration
We can cancel \(m\) from both sides (assuming \(m \neq 0\)):
\[ a = 2g \]
Given \(g = 10 \, \text{m/s}^2\):
\[ a = 2 \cdot 10 = 20 \, \text{m/s}^2 \]

### Final Answer
The instantaneous acceleration of the oil drop after the polarity of the plates is changed is:
\[ \boxed{20 \, \text{m/s}^2} \]