Two cahrges of `50muC` and `100muC` are separated by a distance of `0.6m` .The intensity of electric filed at a point midway between them is

Options

A. `50xx10^(6)v//M`
B. `5xx10^(6)M//V`
C. `10xx10^(6)V//m`
D. `10XX10^(-6)V//M`

Explanation

To find the intensity of the electric field at a point midway between two charges of 50 µC and 100 µC separated by a distance of 0.6 m, we can follow these steps:

### Step 1: Identify the Charges and Distance
We have two charges:
- \( Q_1 = 50 \, \mu C = 50 \times 10^{-6} \, C \)
- \( Q_2 = 100 \, \mu C = 100 \times 10^{-6} \, C \)

The distance between the charges is \( d = 0.6 \, m \). The point of interest is midway between the two charges, which is at a distance of \( r = \frac{d}{2} = 0.3 \, m \) 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:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) is Coulomb's constant.

#### Electric Field due to \( Q_2 \) (100 µC):
\[ E_2 = \frac{k \cdot |Q_2|}{r^2} = \frac{9 \times 10^9 \cdot 100 \times 10^{-6}}{(0.3)^2} \]
Calculating \( (0.3)^2 = 0.09 \):
\[ E_2 = \frac{9 \times 10^9 \cdot 100 \times 10^{-6}}{0.09} = \frac{9 \times 10^9 \cdot 10^{-4}}{0.09} \]
\[ E_2 = 10^7 \, V/m \]

#### Electric Field due to \( Q_1 \) (50 µC):
\[ E_1 = \frac{k \cdot |Q_1|}{r^2} = \frac{9 \times 10^9 \cdot 50 \times 10^{-6}}{(0.3)^2} \]
Using the same \( (0.3)^2 = 0.09 \):
\[ E_1 = \frac{9 \times 10^9 \cdot 50 \times 10^{-6}}{0.09} = \frac{9 \times 10^9 \cdot 5 \times 10^{-5}}{0.09} \]
\[ E_1 = 5 \times 10^6 \, V/m \]

### Step 3: Determine the Direction of the Electric Fields
- The electric field \( E_2 \) due to \( Q_2 \) (100 µC) will point away from \( Q_2 \) towards the midpoint.
- The electric field \( E_1 \) due to \( Q_1 \) (50 µC) will point towards \( Q_1 \).

### Step 4: Calculate the Net Electric Field
Since \( E_2 \) is greater than \( E_1 \) and they are in opposite directions, the net electric field \( E_{net} \) at the midpoint is:
\[ E_{net} = E_2 - E_1 \]
Substituting the values:
\[ E_{net} = 10^7 \, V/m - 5 \times 10^6 \, V/m = 5 \times 10^6 \, V/m \]

### Final Answer
The intensity of the electric field at the point midway between the charges is:
\[ \boxed{5 \times 10^6 \, V/m} \]