Two equal and opposite charges of magnitude `0.2 mu C` are 15 cm apart, the magnitude and direction of the resultant electric intensity at a point midway between the charges is

Options

A. `6.4 xx 10^(5)N//C` towards -ve charge
B. `6.4 xx 10^(5)N//C` towards +ve charge
C. zero
D. infinity

Answer

To find the magnitude and direction of the resultant electric intensity at a point midway between two equal and opposite charges of magnitude \(0.2 \, \mu C\) that are \(15 \, cm\) apart, we can follow these steps:

### Step 1: Understand the setup
We have two charges:
- Charge \(Q_1 = +0.2 \, \mu C = +0.2 \times 10^{-6} \, C\)
- Charge \(Q_2 = -0.2 \, \mu C = -0.2 \times 10^{-6} \, C\)

The distance between the charges is \(d = 15 \, cm = 0.15 \, m\). The point of interest is midway between the two charges, which means it is located at a distance of \(r = \frac{d}{2} = \frac{0.15}{2} = 0.075 \, m\) 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 \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) is Coulomb's constant.

#### Electric field due to \(Q_1\) (positive charge):
\[ E_1 = \frac{k \cdot |Q_1|}{r^2} = \frac{9 \times 10^9 \cdot 0.2 \times 10^{-6}}{(0.075)^2} \]

#### Electric field due to \(Q_2\) (negative charge):
\[ E_2 = \frac{k \cdot |Q_2|}{r^2} = \frac{9 \times 10^9 \cdot 0.2 \times 10^{-6}}{(0.075)^2} \]

### Step 3: Calculate the total electric field at the midpoint
Since both charges are equal in magnitude but opposite in sign, the magnitudes of the electric fields \(E_1\) and \(E_2\) will be the same, but their directions will be opposite.

At the midpoint:
- The electric field due to \(Q_1\) (positive charge) will point away from \(Q_1\) towards the left.
- The electric field due to \(Q_2\) (negative charge) will point towards \(Q_2\) (also towards the left).

Thus, the resultant electric field \(E_{net}\) at the midpoint is:
\[ E_{net} = E_1 + E_2 = 2E_1 \]

### Step 4: Substitute and calculate
Substituting the values:
\[ E_{net} = 2 \cdot \frac{9 \times 10^9 \cdot 0.2 \times 10^{-6}}{(0.075)^2} \]

Calculating \(E_{net}\):
\[ E_{net} = 2 \cdot \frac{9 \times 10^9 \cdot 0.2 \times 10^{-6}}{0.005625} = 6.4 \times 10^5 \, \text{N/C} \]

### Step 5: Determine the direction
The direction of the resultant electric field is towards the positive charge \(Q_1\) because the electric field lines point away from positive charges.

### Final Answer
The magnitude of the resultant electric intensity at the midpoint is \(6.4 \times 10^5 \, \text{N/C}\) and the direction is towards the positive charge.

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