THREE EQUAL CHARGES, 2.0*10 TO POWER OF -6 EACH, ARE HELD FIXED AT THE THREE COMERS OF AN EQUILATERAL TRIANGLE OF SIDE 5 CM. FIND THE COULOMB FORSE EXPERIENCED BY ONE OF THE CHARGES DUE TO THE REST TWO.

To find the Coulomb force experienced by one of the charges due to the other two in an equilateral triangle configuration, we can use Coulomb's law. This law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Let's break this down step by step.

Understanding the Setup

We have three equal charges, each with a magnitude of \(2.0 \times 10^{-6}\) C, positioned at the corners of an equilateral triangle with each side measuring 5 cm (or 0.05 m). We will calculate the force experienced by one charge due to the other two charges.

Coulomb's Law Formula

The formula for Coulomb's law is given by:

F = k * |q1 * q2| / r²

Where:

• F is the force between the charges.
• k is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).
• q1 and q2 are the magnitudes of the charges.
• r is the distance between the charges.

Calculating the Force Between Two Charges

Let's denote the charges as \(q_A\), \(q_B\), and \(q_C\). We will calculate the force on charge \(q_A\) due to charges \(q_B\) and \(q_C\) separately, and then combine these forces vectorially.

Force Between \(q_A\) and \(q_B\)

Using Coulomb's law:

F_{AB} = k * |q_A * q_B| / r^2

Substituting the values:

F_{AB} = (8.99 \times 10^9) * (2.0 \times 10^{-6})^2 / (0.05)^2

Calculating this gives:

F_{AB} = (8.99 \times 10^9) * (4.0 \times 10^{-12}) / (0.0025)

Which simplifies to:

F_{AB} = (8.99 \times 10^9) * (1.6 \times 10^{-9}) = 14.384 \, \text{N}

Force Between \(q_A\) and \(q_C\)

Since the charges are equal and symmetrically placed, the force \(F_{AC}\) will be the same as \(F_{AB}\):

F_{AC} = 14.384 \, \text{N}

Combining the Forces

Now, we need to find the resultant force on \(q_A\) due to both \(q_B\) and \(q_C\). The forces \(F_{AB}\) and \(F_{AC}\) are at an angle of 60 degrees to each other because of the equilateral triangle configuration.

Using Vector Addition

The resultant force \(F_R\) can be calculated using the formula:

F_R = √(F_{AB}² + F_{AC}² + 2 * F_{AB} * F_{AC} * cos(θ))

Where \(θ = 60°\). The cosine of 60 degrees is 0.5. Plugging in the values:

F_R = √(14.384² + 14.384² + 2 * 14.384 * 14.384 * 0.5)

This simplifies to:

F_R = √(2 * 14.384² + 14.384²) = √(3 * 14.384²)

Calculating this gives:

F_R = 14.384 * √3 ≈ 24.84 \, \text{N}

Final Result

The total Coulomb force experienced by one of the charges due to the other two is approximately 24.84 N. This force acts outward from the center of the triangle, directed away from the other two charges.