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. `4.5xx10^(6)N//C` directed towards `+5 mu C`
B. `4.5xx10^(6)N//C` directed towards `+10 mu C`
C. `13.5xx10^(6)N//C` directed towards `+5 mu C`
D. `13.5xx10^(6)N//C` directed towards `+10 mu C`

Explanation

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

### Step 1: Understand the Setup
- We have two charges:
- Charge \( q_1 = +5 \mu C = 5 \times 10^{-6} C \)
- Charge \( q_2 = +10 \mu C = 10 \times 10^{-6} C \)
- The distance between the charges is \( 20 \, cm = 0.2 \, m \).
- The midpoint between the two charges is \( 10 \, cm = 0.1 \, m \) from each charge.

### 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, and \( r \) is the distance from the charge to the point where the electric field is being calculated.

#### Electric Field Due to \( q_1 \) (5 μC):
- Distance \( r = 0.1 \, m \)
\[ E_1 = \frac{9 \times 10^9 \cdot 5 \times 10^{-6}}{(0.1)^2} \]
\[ E_1 = \frac{9 \times 10^9 \cdot 5 \times 10^{-6}}{0.01} \]
\[ E_1 = 9 \times 5 \times 10^3 = 45 \times 10^3 \, N/C = 4.5 \times 10^6 \, N/C \]

#### Electric Field Due to \( q_2 \) (10 μC):
- Distance \( r = 0.1 \, m \)
\[ E_2 = \frac{9 \times 10^9 \cdot 10 \times 10^{-6}}{(0.1)^2} \]
\[ E_2 = \frac{9 \times 10^9 \cdot 10 \times 10^{-6}}{0.01} \]
\[ E_2 = 9 \times 10 \times 10^3 = 90 \times 10^3 \, N/C = 9 \times 10^6 \, N/C \]

### Step 3: Determine the Direction of the Electric Fields
- The electric field due to both charges will point away from the charges since they are both positive.
- Therefore, at the midpoint:
- The electric field \( E_1 \) due to \( +5 \mu C \) will point to the right.
- The electric field \( E_2 \) due to \( +10 \mu C \) will point to the left.

### Step 4: Calculate the Net Electric Field
Since \( E_2 \) is greater than \( E_1 \), the net electric field \( E_{net} \) will be:
\[ E_{net} = E_2 - E_1 \]
\[ E_{net} = 9 \times 10^6 - 4.5 \times 10^6 = 4.5 \times 10^6 \, N/C \]

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

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