A and B are two points separately by a distance 5cm. Two charges `10 mu C and 20mu C` are placed at A and B. The resultant electric intensity at a point P outside the charges at a distance 5cm from `10 mu C` is

Options

A. `54 xx 10^(6) N//C` away from `10 mu C`
B. `56 xx 10^(6)N//C` towards `10 mu C`
C. `9 xx 10^(6) N//C` away from `10 mu C`
D. zero

Answer

To solve the problem, we need to find the resultant electric field intensity at point P, which is located 5 cm from the charge of 10 µC (microcoulombs) at point A. The charge at point B is 20 µC and is located 5 cm away from point A.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- Charge at point A (Q1) = 10 µC = \(10 \times 10^{-6}\) C
- Charge at point B (Q2) = 20 µC = \(20 \times 10^{-6}\) C
- Distance between A and B = 5 cm = \(0.05\) m
- Distance from point P to charge A = 5 cm = \(0.05\) m
- Distance from point P to charge B = Distance from A to B + Distance from A to P = \(0.05 + 0.05 = 0.10\) m

2. **Calculate the Electric Field due to Charge A (E1):**
The electric field (E) due to a point charge is given by the formula:
\[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r^2} \]
Where:
- \( \epsilon_0 \) (permittivity of free space) = \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( Q \) = charge
- \( r \) = distance from the charge

For charge A:
\[ E_1 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{10 \times 10^{-6}}{(0.05)^2} \]
\[ E_1 = \frac{9 \times 10^9}{(0.0025)} \cdot 10 \times 10^{-6} \]
\[ E_1 = 9 \times 10^9 \cdot 4 \times 10^3 = 36 \times 10^6 \, \text{N/C} \]

3. **Calculate the Electric Field due to Charge B (E2):**
For charge B:
\[ E_2 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{20 \times 10^{-6}}{(0.10)^2} \]
\[ E_2 = \frac{9 \times 10^9}{(0.01)} \cdot 20 \times 10^{-6} \]
\[ E_2 = 9 \times 10^9 \cdot 2 \times 10^6 = 18 \times 10^6 \, \text{N/C} \]

4. **Determine the Direction of the Electric Fields:**
- Both electric fields \( E_1 \) and \( E_2 \) are directed towards the positive charges (since they are positive), which means they are directed towards point B.

5. **Calculate the Resultant Electric Field (E):**
Since both fields are in the same direction, we can add them:
\[ E = E_1 + E_2 = 36 \times 10^6 + 18 \times 10^6 = 54 \times 10^6 \, \text{N/C} \]

6. **Final Result:**
The resultant electric intensity at point P is \( 54 \times 10^6 \, \text{N/C} \) directed away from the charge at A.