Two point charges of `1muC and -1muC` are separated by a distance of 100 Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be

Options

A. `9NC^(-1)`
B. `0.9NC^(-1)`
C. `90NC^(-1)`
D. `0.09NC^(-1)`

Explanation

To find the electric field at point P due to the two point charges of \(1 \mu C\) and \(-1 \mu C\) separated by a distance of \(100 \, \text{Å}\), we can follow these steps:

### Step 1: Understand the Configuration
We have two point charges:
- \(Q_1 = +1 \mu C = 1 \times 10^{-6} C\)
- \(Q_2 = -1 \mu C = -1 \times 10^{-6} C\)

The distance between the charges is \(100 \, \text{Å} = 100 \times 10^{-10} \, \text{m} = 1 \times 10^{-8} \, \text{m}\).

Point P is located at a distance of \(10 \, \text{cm} = 0.1 \, \text{m}\) from the midpoint of the line joining the two charges, on the perpendicular bisector.

### Step 2: Calculate the Distance from Each Charge to Point P
The distance from the midpoint to each charge is:
\[ d = \frac{100 \, \text{Å}}{2} = 50 \, \text{Å} = 50 \times 10^{-10} \, \text{m} = 5 \times 10^{-9} \, \text{m} \]

Using the Pythagorean theorem, the distance from each charge to point P (let's call it \(R\)) is:
\[ R = \sqrt{(0.1)^2 + (5 \times 10^{-9})^2} \approx 0.1 \, \text{m} \quad (\text{since } 5 \times 10^{-9} \text{ is negligible compared to } 0.1) \]

### Step 3: Calculate the Electric Field Due to Each Charge
The electric field \(E\) due to a point charge is given by:
\[ E = \frac{k \cdot |Q|}{R^2} \]
where \(k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2\).

For charge \(Q_1\):
\[ E_1 = \frac{9 \times 10^9 \cdot 1 \times 10^{-6}}{(0.1)^2} = \frac{9 \times 10^3}{0.01} = 9 \times 10^5 \, \text{N/C} \]

For charge \(Q_2\):
\[ E_2 = \frac{9 \times 10^9 \cdot 1 \times 10^{-6}}{(0.1)^2} = 9 \times 10^5 \, \text{N/C} \]

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

Since point P is on the perpendicular bisector, both electric fields will have the same magnitude but opposite directions along the line connecting the charges.

### Step 5: Calculate the Net Electric Field at Point P
Since \(E_1\) and \(E_2\) are equal in magnitude and opposite in direction, they will cancel each other out:
\[ E_{\text{net}} = E_1 - E_2 = 9 \times 10^5 - 9 \times 10^5 = 0 \, \text{N/C} \]

### Step 6: Conclusion
The electric field at point P is:
\[ E = 0 \, \text{N/C} \]