Two charges `+5mu C` and `+10 mu C` are placed `20 cm` apart. The net electric field at the mid-point between the two charges is

Options

A. `4.5 xx 10^(6) NC^(-1)` directed towards `+ 5muC`
B. `4.5 xx 10^(6) NC^(-1)` directed towards `+ 10 muC`
C. `13.5 xx 10^(6) NC^(-1)` directed towards `+ 5 muC`
D. `13.5 xx 10^(6) NC^(-1)` directed towards `+ 10 muC`

Answer

To find the net electric field at the midpoint between two charges \( +5 \mu C \) and \( +10 \mu C \) placed \( 20 \, cm \) apart, we can follow these steps:

### Step 1: Identify the positions of the charges
Let’s denote the charges as:
- Charge \( Q_1 = +5 \mu C \) located at point A.
- Charge \( Q_2 = +10 \mu C \) located at point B.

The distance between the two charges is \( 20 \, cm \), which means the midpoint (point M) is \( 10 \, cm \) from both charges.

### 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 field is being calculated.

#### Electric Field due to \( Q_1 \) at M:
- Distance \( r_1 = 10 \, cm = 0.1 \, m \)
- Charge \( Q_1 = +5 \mu C = 5 \times 10^{-6} \, C \)

Calculating \( E_1 \):
\[ E_1 = \frac{9 \times 10^9 \cdot 5 \times 10^{-6}}{(0.1)^2} = \frac{9 \times 10^9 \cdot 5 \times 10^{-6}}{0.01} = 4.5 \times 10^6 \, N/C \]
This electric field is directed away from \( Q_1 \) (to the right, along the positive x-axis).

#### Electric Field due to \( Q_2 \) at M:
- Distance \( r_2 = 10 \, cm = 0.1 \, m \)
- Charge \( Q_2 = +10 \mu C = 10 \times 10^{-6} \, C \)

Calculating \( E_2 \):
\[ E_2 = \frac{9 \times 10^9 \cdot 10 \times 10^{-6}}{(0.1)^2} = \frac{9 \times 10^9 \cdot 10 \times 10^{-6}}{0.01} = 9 \times 10^6 \, N/C \]
This electric field is also directed away from \( Q_2 \) (to the left, along the negative x-axis).

### Step 3: Determine the direction of the net electric field
Since both charges are positive, the electric field due to \( Q_1 \) (to the right) and \( Q_2 \) (to the left) will oppose each other. Therefore, the net electric field will be directed towards the charge with the smaller magnitude, which is \( Q_1 \).

### Step 4: Calculate the net electric field
The net electric field \( E_{net} \) at the midpoint is given by:
\[ E_{net} = E_2 - E_1 \]
Substituting the values:
\[ E_{net} = 9 \times 10^6 - 4.5 \times 10^6 = 4.5 \times 10^6 \, N/C \]
This net electric field is directed towards \( Q_1 \).

### Final Answer
The net electric field at the midpoint between the two charges is:
\[ E_{net} = 4.5 \times 10^6 \, N/C \, \text{(towards } +5 \mu C\text{)} \]