Two charges 10nC & 20nC are laced at the corners of hypotenuse BC of a right angled triangle ABC of side AB = 3m \angle AC=4m Find (a) magnitude of force acting on corner A of right angled triangle ABC if a charge +2nC is placed at corner A (b) direction of resultant force at corner A

The problem involves the calculation of electric forces between charges, which is governed by Coulomb's Law. The law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

(a) Magnitude of force acting on corner A:

First, we calculate the force between the charges at A and B. Let's denote this force as F_AB. According to Coulomb's Law, F_AB = k * |q_A * q_B| / r_AB^2, where k is Coulomb's constant (approximately 9 * 10^9 N m^2/C^2), q_A and q_B are the charges at A and B (2 nC and 10 nC, respectively), and r_AB is the distance between A and B (3 m). Note that we need to convert the charges from nC to C by multiplying by 10^-9.

F_AB = 9 * 10^9 N m^2/C^2 * |2 * 10^-9 C * 10 * 10^-9 C| / (3 m)^2 = 0.02 N.

Similarly, we calculate the force between the charges at A and C. Let's denote this force as F_AC. Here, q_A and q_C are the charges at A and C (2 nC and 20 nC, respectively), and r_AC is the distance between A and C (4 m).

F_AC = 9 * 10^9 N m^2/C^2 * |2 * 10^-9 C * 20 * 10^-9 C| / (4 m)^2 = 0.045 N.

The total force at A is the vector sum of F_AB and F_AC. Since the triangle is right-angled, these forces are perpendicular to each other, so we can use the Pythagorean theorem to find the magnitude of the total force: F_A = sqrt(F_AB^2 + F_AC^2) = sqrt((0.02 N)^2 + (0.045 N)^2) = 0.05 N.

(b) Direction of resultant force at corner A:

The direction of the resultant force can be found using trigonometry. The angle θ between the resultant force and the force F_AB (or the AB side of the triangle) can be found from the tangent of the angle: tan(θ) = F_AC / F_AB.

θ = arctan(F_AC / F_AB) = arctan(0.045 N / 0.02 N) = 66.8 degrees.

So, the direction of the resultant force at corner A is 66.8 degrees from the AB side of the triangle, towards the AC side.