ABC is a equilateral triangle of side 0.1m . Charges are placed at the vertices A and B respectively.Calculate the resultant electric intensity at a point C.

Answer 1

To calculate the resultant electric intensity at point C, we need to find the electric intensities due to charges placed at A and B, and then find their vector sum.

Let's assume the charge at A is q1 and at B is q2.

The electric intensity E due to a point charge q at a distance r from the point can be given by Coulomb's law:

E = k * q / r^2

where k is the electrostatic constant, approximately 8.9875 * 10^9 N m^2 C^(-2).

Now let's calculate the electric intensities due to charges q1 and q2 at point C.

Since ABC is an equilateral triangle with side 0.1m, the distance between charges q1 and C, and charges q2 and C are both equal to 0.1m.

E1 due to charge q1 at point C:

E1 = k * q1 / (0.1)^2

E2 due to charge q2 at point C:

E2 = k * q2 / (0.1)^2

Now to find the resultant electric intensity at point C, we need to find the vector sum of E1 and E2.

Since the electric field is a vector quantity, we need to consider both magnitude and direction.

Since ABC is an equilateral triangle, the angle between E1 and E2 is 60 degrees. Since the charges are at vertices A and B, the electric fields E1 and E2 are along the lines AC and BC, respectively, and their directions are away from A and B towards C. As a result, we can apply the law of cosines to find the resultant electric field at C:

E_resultant^2 = E1^2 + E2^2 + 2 * E1 * E2 * cos(120°)

Since cos(120°) = -1/2, we get:

E_resultant^2 = E1^2 + E2^2 - E1 * E2

Substituting the values of E1 and E2 in terms of q1 and q2, we get:

E_resultant^2 = (k^2 * q1^2 / (0.1)^4) + (k^2 * q2^2 / (0.1)^4) - (k^2 * q1 * q2 / (0.1)^4)

E_resultant = (k / (0.1)^2) * sqrt(q1^2 + q2^2 - q1 * q2)

Now, we can't proceed further without knowing the values of q1 and q2. Once we know the values of the charges, we can plug them into the above equation to find the resultant electric intensity at point C.