Four charges of +q, +q +q and +q are placed at the corners A, B, C and D of s square. The resultant force on the charge at D

Options

A. `(q^(2))/(8pi in_(0)a^(2)) (1 + 2 sqrt2)`
B. `(2q^(2))/(pi in_(0)a^(2))`
C. `(q^(2))/(8pi in_(0)a^(2)) x 2 sqrt2`
D. zero

Answer

To find the resultant force on the charge at point D due to the other three charges at points A, B, and C, we can follow these steps:

### Step 1: Identify the Charges and Their Positions
We have four charges of +q placed at the corners of a square ABCD. The positions of the charges are:
- Charge at A: +q
- Charge at B: +q
- Charge at C: +q
- Charge at D: +q (the charge we are analyzing)

### Step 2: Calculate the Forces Acting on Charge D
The forces acting on charge D due to the other charges can be calculated using Coulomb's Law, which states that the force between two charges is given by:

\[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

#### Force on D due to A (F_AD):
The distance from A to D is equal to the side of the square (a):
\[ F_{AD} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \]

#### Force on D due to B (F_BD):
The distance from B to D is also equal to the side of the square (a):
\[ F_{BD} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \]

#### Force on D due to C (F_CD):
The distance from C to D is equal to the diagonal of the square, which is \( \sqrt{2}a \):
\[ F_{CD} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{( \sqrt{2} a )^2} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{2a^2} \]

### Step 3: Determine the Directions of the Forces
- The force \( F_{AD} \) acts directly towards A (to the left).
- The force \( F_{BD} \) acts directly towards B (upwards).
- The force \( F_{CD} \) acts directly towards C (diagonally downwards).

### Step 4: Resolve Forces into Components
Since \( F_{AD} \) and \( F_{BD} \) are perpendicular to each other, we can use the Pythagorean theorem to find the resultant of these two forces.

Let:
- \( F_{AD} = F_1 = \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \)
- \( F_{BD} = F_2 = \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \)

The resultant force \( F_{AB} \) from A and B is:
\[ F_{AB} = \sqrt{F_1^2 + F_2^2} = \sqrt{\left(\frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2}\right)^2 + \left(\frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2}\right)^2} \]
\[ = \sqrt{2 \left(\frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2}\right)^2} \]
\[ = \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \sqrt{2} \]

### Step 5: Add the Force from C
Now, we need to add the force from C, which is acting diagonally towards D:
\[ F_{CD} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{2a^2} \]

### Step 6: Calculate the Total Resultant Force on D
The total force \( F_{net} \) on charge D is the vector sum of \( F_{AB} \) and \( F_{CD} \):
\[ F_{net} = F_{AB} + F_{CD} \]
Substituting the values:
\[ F_{net} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \sqrt{2} + \frac{1}{4\pi \epsilon_0} \frac{q^2}{2a^2} \]
Factoring out \( \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \):
\[ F_{net} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) \]

### Final Result
Thus, the resultant force on the charge at D is:
\[ F_{net} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2} \left( \frac{1}{2} + \sqrt{2} \right) \]