Each of +3micro C are placed at three corners of a square whose diagonal is 6 m long find feild intensity at the point if intersection of diagonal
you only need to find the field intensity fro the "middle" corner
... the two diagonally opposite corners will cancel each other
To find the electric field intensity at the point of intersection of the diagonal, we can use the principle of superposition. The electric field created by each charge will contribute to the total electric field at that point.
Given that each charge is +3 micro C (3 x 10^-6 C) and they are placed at the three corners of a square, the electric field created by each charge can be calculated using Coulomb's law.
Coulomb's law states that the electric field created by a point charge is given by the equation:
E = k * (Q / r^2)
Where:
E is the electric field intensity,
k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2),
Q is the charge, and
r is the distance between the charge and the point where we want to calculate the electric field.
In this case, the distance between each charge and the point of interest (the intersection of the diagonal) is the length of the diagonal, which is 6 m.
Let's calculate the electric field created by each charge:
E1 = k * (Q1 / r^2) = (8.99 x 10^9 Nm^2/C^2) * (3 x 10^-6 C) / (6 m)^2
E1 = (8.99 x 10^9 Nm^2/C^2) * (3 x 10^-6 C) / 36 m^2
E1 = 0.7491667 x 10^3 N/C
Since the square has four corners and we only have three charges, we will need to consider the fields created by the other two charges as well. However, due to the symmetry of the problem, we can determine that the field created by the other two charges will be equal in magnitude and opposite in direction to E1.
Therefore, the total electric field at the point of intersection of the diagonal is:
E_total = E1 + (-E1) + (-E1) = -E1
The electric field is negative because the charges are all positive, and the field points in the opposite direction.
So, the electric field intensity at the point of intersection of the diagonal is approximately -0.749 x 10^3 N/C.