Two parallel plates separated by a distance of 5 mm are kept at a potential difference of 50 V. A particle of mass `10^(-15)` and charge `10^(-11)` C enters in it with a velocity `10^(7)m//s`. The acceleration of the particle will be

Options

A. `10^(8)m//s^(2)`
B. `5 xx 10^(5) m//s^(2)`
C. `10^(5) m//s^(2)`
D. `2 xx 10^(3) m//s^(2)`

Answer

To find the acceleration of a charged particle entering between two parallel plates, we can follow these steps:

### Step 1: Understand the Problem
We have two parallel plates separated by a distance \( d = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \) with a potential difference \( V = 50 \, \text{V} \). A particle with mass \( m = 10^{-15} \, \text{kg} \) and charge \( q = 10^{-11} \, \text{C} \) enters the region between the plates with an initial velocity \( v = 10^{7} \, \text{m/s} \). We need to find the acceleration \( a \) of the particle.

### Step 2: Calculate the Electric Field
The electric field \( E \) between the plates can be calculated using the formula:
\[ E = \frac{V}{d} \]
Substituting the values:
\[ E = \frac{50 \, \text{V}}{5 \times 10^{-3} \, \text{m}} = \frac{50}{0.005} = 10000 \, \text{V/m} \]

### Step 3: Calculate the Electric Force
The electric force \( F \) acting on the particle is given by:
\[ F = q \cdot E \]
Substituting the values:
\[ F = (10^{-11} \, \text{C}) \cdot (10000 \, \text{V/m}) = 10^{-11} \cdot 10^{4} = 10^{-7} \, \text{N} \]

### Step 4: Apply Newton's Second Law
According to Newton's second law, the acceleration \( a \) can be calculated using the formula:
\[ F = m \cdot a \implies a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{10^{-7} \, \text{N}}{10^{-15} \, \text{kg}} = 10^{8} \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle is:
\[ a = 10^{8} \, \text{m/s}^2 \]

---