Two charges `10 muC and - 10 muC` are placed at points `A. and B` separated by a distance of 10 cm. Find the electric field at a point P on the perpendicular bisector of `AB` at. a distance of 12 cm from its middle point.

Options

A. `4.1 xx 10^6 N//C`
B. `2.2 xx 10^6 N//C`
C. `4.1 xx 10^5 N//C`
D. `2.2 xx 10^5 N//C`

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Configuration
We have two charges:
- Charge \( Q_1 = +10 \, \mu C = 10 \times 10^{-6} \, C \) at point A.
- Charge \( Q_2 = -10 \, \mu C = -10 \times 10^{-6} \, C \) at point B.
- The distance between A and B is \( d = 10 \, cm = 0.1 \, m \).

Point P is located on the perpendicular bisector of AB, at a distance of \( 12 \, cm = 0.12 \, m \) from the midpoint of AB.

### Step 2: Calculate the Distance from Each Charge to Point P
The distance from the midpoint of AB to point P is \( 12 \, cm \). The distance from each charge to point P can be calculated using the Pythagorean theorem.

Let \( x \) be the distance from the midpoint to point P (which is \( 12 \, cm \)) and \( \frac{d}{2} \) be the distance from the midpoint to either charge (which is \( 5 \, cm \)):
\[ R = \sqrt{\left(\frac{d}{2}\right)^2 + x^2} = \sqrt{(5 \, cm)^2 + (12 \, cm)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm = 0.13 \, m \]

### Step 3: Calculate the Electric Field Due to Each Charge
The electric field \( E \) due to a point charge is given by:
\[ E = \frac{k |Q|}{R^2} \]
where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) is Coulomb's constant.

For both charges:
1. For \( Q_1 \):
\[ E_1 = \frac{9 \times 10^9 \times 10 \times 10^{-6}}{(0.13)^2} = \frac{9 \times 10^9 \times 10 \times 10^{-6}}{0.0169} \approx 5.3 \times 10^6 \, N/C \]

2. For \( Q_2 \):
\[ E_2 = \frac{9 \times 10^9 \times 10 \times 10^{-6}}{(0.13)^2} = \frac{9 \times 10^9 \times 10 \times 10^{-6}}{0.0169} \approx 5.3 \times 10^6 \, N/C \]

### Step 4: Determine the Direction of the Electric Fields
- The electric field \( E_1 \) due to the positive charge \( Q_1 \) points away from the charge (towards point P).
- The electric field \( E_2 \) due to the negative charge \( Q_2 \) points towards the charge (also towards point P).

### Step 5: Calculate the Net Electric Field
Since both electric fields point in the same direction (towards point P), the net electric field \( E_{net} \) is the sum of the magnitudes of the two electric fields:
\[ E_{net} = E_1 + E_2 = 5.3 \times 10^6 + 5.3 \times 10^6 = 10.6 \times 10^6 \, N/C \]

### Final Answer
The electric field at point P is:
\[ E_{net} \approx 10.6 \times 10^6 \, N/C \]