Two positive charges of 1mu C` and `2 mu C` are placed 1 metre apart. The value of electric field in N/C at the middle point of the line joining the charge will be :

Options

A. `10.8 xx 10^(4)`
B. `3.6 xx 10^(4)`
C. `1.8 xx 10^(4)`
D. `5.4 xx 10^(4)`

Answer

To find the electric field at the midpoint between two positive charges of \(1 \mu C\) and \(2 \mu C\) placed 1 meter apart, we can follow these steps:

### Step 1: Identify the charges and their positions
Let:
- Charge \(Q_1 = 1 \mu C = 1 \times 10^{-6} C\) located at point A (0 m).
- Charge \(Q_2 = 2 \mu C = 2 \times 10^{-6} C\) located at point B (1 m).
- The midpoint between the two charges is at \(0.5 m\).

### Step 2: Calculate the electric field due to each charge at the midpoint
The electric field \(E\) due to a point charge is given by the formula:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where:
- \(k = 9 \times 10^9 \, N \cdot m^2/C^2\) (Coulomb's constant),
- \(Q\) is the charge,
- \(r\) is the distance from the charge to the point where the electric field is being calculated.

#### Electric Field due to \(Q_1\) at midpoint:
- Distance from \(Q_1\) to midpoint = \(0.5 m\)
\[ E_1 = \frac{k \cdot |Q_1|}{(0.5)^2} = \frac{9 \times 10^9 \cdot 1 \times 10^{-6}}{(0.5)^2} = \frac{9 \times 10^9 \cdot 1 \times 10^{-6}}{0.25} = 36 \times 10^3 \, N/C = 36000 \, N/C \]
The direction of \(E_1\) is away from \(Q_1\) (to the right).

#### Electric Field due to \(Q_2\) at midpoint:
- Distance from \(Q_2\) to midpoint = \(0.5 m\)
\[ E_2 = \frac{k \cdot |Q_2|}{(0.5)^2} = \frac{9 \times 10^9 \cdot 2 \times 10^{-6}}{(0.5)^2} = \frac{9 \times 10^9 \cdot 2 \times 10^{-6}}{0.25} = 72 \times 10^3 \, N/C = 72000 \, N/C \]
The direction of \(E_2\) is away from \(Q_2\) (to the left).

### Step 3: Determine the net electric field at the midpoint
Since both charges are positive, the electric fields due to each charge will point in opposite directions:
- \(E_1\) points to the right (positive direction).
- \(E_2\) points to the left (negative direction).

Thus, the net electric field \(E_{net}\) at the midpoint is:
\[ E_{net} = E_2 - E_1 = 72000 \, N/C - 36000 \, N/C = 36000 \, N/C \]
The direction of the net electric field is towards \(Q_2\) (to the left).

### Final Answer
The value of the electric field at the midpoint of the line joining the charges is \(36000 \, N/C\) directed towards the \(2 \mu C\) charge.

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