Three point charges q are placed at three vertices of an equilateral triangle of side a. Find magnitude of electric force on any charge due to the other two.

Options

A. `(sqrt3/(4piepsilon_0))(q/a)^2`
B. `(sqrt2/(4piepsilon_0))(q/a)^2`
C. `(sqrt3/(4piepsilon_0))(q/(2a))^2`
D. `(sqrt2/(4piepsilon_0))((2q)/a)^2`

Explanation

To find the magnitude of the electric force on any charge due to the other two charges placed at the vertices of an equilateral triangle, we can follow these steps:

### Step 1: Understand the Configuration
Consider three point charges, each with charge \( q \), placed at the vertices \( A \), \( B \), and \( C \) of an equilateral triangle with side length \( a \). We will calculate the net electric force acting on charge \( q \) at vertex \( A \) due to the other two charges at vertices \( B \) and \( C \).

### Step 2: Calculate the Force Between Charges
According to Coulomb's Law, the electric force \( F \) between two point charges is given by:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:
- \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

In our case, the distance \( r \) between any two charges is \( a \). Therefore, the force between any two charges \( q \) at distance \( a \) is:
\[ F_{AB} = F_{AC} = k \frac{q^2}{a^2} \]

### Step 3: Determine the Direction of Forces
Since we are assuming that all charges have the same sign, the forces will be repulsive. Thus:
- The force \( F_{AB} \) acts on charge \( A \) due to charge \( B \) and points away from \( B \) towards \( A \).
- The force \( F_{AC} \) acts on charge \( A \) due to charge \( C \) and points away from \( C \) towards \( A \).

### Step 4: Resolve Forces into Components
To find the net force on charge \( A \), we need to resolve the forces \( F_{AB} \) and \( F_{AC} \) into their components. The angle between the lines connecting \( A \) to \( B \) and \( A \) to \( C \) is \( 60^\circ \).

The horizontal component (x-direction) and vertical component (y-direction) of the forces can be calculated as follows:
- The x-component of \( F_{AB} \):
\[ F_{ABx} = F_{AB} \cos(60^\circ) = F_{AB} \cdot \frac{1}{2} = \frac{1}{2} k \frac{q^2}{a^2} \]
- The y-component of \( F_{AB} \):
\[ F_{ABy} = F_{AB} \sin(60^\circ) = F_{AB} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} k \frac{q^2}{a^2} \]

Similarly, for \( F_{AC} \):
- The x-component of \( F_{AC} \):
\[ F_{ACx} = F_{AC} \cos(60^\circ) = \frac{1}{2} k \frac{q^2}{a^2} \]
- The y-component of \( F_{AC} \):
\[ F_{ACy} = F_{AC} \sin(60^\circ) = \frac{\sqrt{3}}{2} k \frac{q^2}{a^2} \]

### Step 5: Calculate the Net Force Components
Now, we can find the net force components acting on charge \( A \):
- The total x-component:
\[ F_{Ax} = F_{ABx} + F_{ACx} = \frac{1}{2} k \frac{q^2}{a^2} + \frac{1}{2} k \frac{q^2}{a^2} = k \frac{q^2}{a^2} \]
- The total y-component:
\[ F_{Ay} = F_{ABy} - F_{ACy} = \frac{\sqrt{3}}{2} k \frac{q^2}{a^2} + \frac{\sqrt{3}}{2} k \frac{q^2}{a^2} = \sqrt{3} k \frac{q^2}{a^2} \]

### Step 6: Calculate the Magnitude of the Net Force
The magnitude of the net force \( F_A \) on charge \( A \) can be calculated using the Pythagorean theorem:
\[ F_A = \sqrt{(F_{Ax})^2 + (F_{Ay})^2} \]
Substituting the values:
\[ F_A = \sqrt{\left(k \frac{q^2}{a^2}\right)^2 + \left(\sqrt{3} k \frac{q^2}{a^2}\right)^2} \]
\[ F_A = \sqrt{k^2 \frac{q^4}{a^4} + 3 k^2 \frac{q^4}{a^4}} = \sqrt{4 k^2 \frac{q^4}{a^4}} = 2 k \frac{q^2}{a^2} \]

### Final Result
Thus, the magnitude of the electric force on any charge due to the other two charges is:
\[ F = 2 k \frac{q^2}{a^2} \]