Two pont chargees `q_(1) " and " q_(2) " of " 2xx10^(-8) C " and " -2xx10^(-8) C` respectively are placed 0.4 m apart. Calculate the electric field at the centre of the line joining the two charges.

Answer

To solve the problem of finding the electric field at the center of the line joining two point charges \( q_1 \) and \( q_2 \), we can follow these steps:

### Step 1: Identify the Charges and Distance
We have two point charges:
- \( q_1 = 2 \times 10^{-8} \, \text{C} \) (positive charge)
- \( q_2 = -2 \times 10^{-8} \, \text{C} \) (negative charge)

The distance between the two charges is given as \( 0.4 \, \text{m} \). The center of the line joining the two charges is \( 0.2 \, \text{m} \) from each charge.

### Step 2: Calculate the Electric Field due to Each Charge
The electric field \( E \) due to a point charge is given by the formula:
\[ E = \frac{k |q|}{r^2} \]
where:
- \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) (Coulomb's constant)
- \( |q| \) is the magnitude of the charge
- \( r \) is the distance from the charge to the point where we are calculating the electric field

#### Electric Field due to \( q_1 \) at the center:
\[ E_1 = \frac{k |q_1|}{(0.2)^2} = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{0.04} \]
Calculating this:
\[ E_1 = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{0.04} = \frac{18 \times 10^1}{0.04} = 4500 \, \text{N/C} \]

#### Electric Field due to \( q_2 \) at the center:
\[ E_2 = \frac{k |q_2|}{(0.2)^2} = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{0.04} \]
Calculating this:
\[ E_2 = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{0.04} = \frac{18 \times 10^1}{0.04} = 4500 \, \text{N/C} \]

### Step 3: Determine the Direction of Electric Fields
- The electric field \( E_1 \) due to the positive charge \( q_1 \) will be directed away from \( q_1 \).
- The electric field \( E_2 \) due to the negative charge \( q_2 \) will be directed towards \( q_2 \).

Since both electric fields are directed towards the negative charge, they will add up.

### Step 4: Calculate the Net Electric Field
The net electric field \( E_{\text{net}} \) at the center is:
\[ E_{\text{net}} = E_1 + E_2 = 4500 + 4500 = 9000 \, \text{N/C} \]

### Step 5: Direction of the Net Electric Field
The direction of the net electric field will be towards the negative charge \( q_2 \).

### Final Answer
The electric field at the center of the line joining the two charges is:
\[ E_{\text{net}} = 9000 \, \text{N/C} \quad \text{(towards the negative charge)} \]

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