Two electric charges +Q and -Q are arranged at the vertices A and B of an isosceles right angled triangle ABC.The electric intensity at C is

To determine the electric field intensity at point C due to the charges +Q and -Q located at points A and B of an isosceles right triangle, we need to analyze the contributions of each charge separately and then combine them vectorially.

Understanding the Setup

In our scenario, we have two charges: +Q at vertex A and -Q at vertex B of triangle ABC. The triangle is isosceles and right-angled, meaning that the lengths of sides AC and BC are equal, and angle C is 90 degrees. Let's denote the length of sides AC and BC as 'd'. Therefore, the distance between the charges A and B (the hypotenuse) can be calculated using the Pythagorean theorem:

• Distance AB = √(d² + d²) = √(2d²) = d√2

Calculating Electric Field Contributions

The electric field (E) created by a point charge is given by the formula:

E = k * |Q| / r²

where:

• E is the electric field intensity.
• k is Coulomb's constant (approximately 8.99 x 10⁹ N m²/C²).
• r is the distance from the charge to the point where the field is being calculated.

Electric Field at Point C due to Charge +Q

The distance from charge +Q at A to point C is 'd'. Therefore, the electric field intensity at C due to +Q is:

E_A = k * Q / d²

This electric field points away from charge +Q (since it is positive), directed along the line AC.

Electric Field at Point C due to Charge -Q

Similarly, the distance from charge -Q at B to point C is also 'd'. The electric field intensity at C due to -Q is:

E_B = k * |Q| / d²

Combining the Electric Fields

Now, we need to combine the electric fields E_A and E_B. Since both fields are at right angles to each other (E_A along AC and E_B along BC), we can use the Pythagorean theorem to find the resultant electric field intensity at C:

E_C = √(E_A² + E_B²)

Substituting the expressions for E_A and E_B:

E_C = √((k * Q / d²)² + (k * Q / d²)²)

E_C = √(2 * (k * Q / d²)²)

E_C = (k * Q / d²) * √2

Direction of the Resultant Electric Field

The direction of the resultant electric field at point C can be determined by considering the individual contributions. The electric field due to +Q (E_A) points away from A towards C, while the electric field due to -Q (E_B) points towards B from C. The resultant electric field will be directed diagonally, towards the midpoint of line AB, making an angle of 45 degrees with both AC and BC.

Final Expression

In summary, the electric field intensity at point C due to the charges +Q and -Q is:

E_C = (k * Q / d²) * √2

This result illustrates how electric fields from different charges can interact and combine, leading to a net electric field at a specific point in space.