ABC is a triangle with sides AB = 3 m, BC = 4 m and ABC = 90°. Charges of +9 x 10-5 C and -
16 x 10-5 C are placed at corners A and C respectively. Find the magnitude and direction of
electric intensity at corner B.
To find the electric intensity at corner B, we need to first calculate the electric field created by each charge at that point, and then vectorially add the two fields.
The electric field created by a point charge q at a distance r is given by:
E = k * (q / r^2) * u,
where k is the Coulomb's constant (8.99 x 10^9 N m^2/C^2) and u is a unit vector pointing from the charge towards the point of interest.
For the +9 x 10^-5 C charge at A:
E_A = (8.99 x 10^9) * (9 x 10^-5) / (3)^2 = 8 x 10^4 N/C,
and the direction of E_A is along the line AB, pointing towards B.
For the -16 x 10^-5 C charge at C:
E_C = (8.99 x 10^9) * (16 x 10^-5) / (5)^2 = 5.15 x 10^4 N/C,
and the direction of E_C is along the line BC, pointing towards B.
Now, we can find the net electric field at B by vectorially adding E_A and E_C:
E_net = √(E_A^2 + E_C^2) = √((8 x 10^4)^2 + (5.15 x 10^4)^2) = 9.46 x 10^4 N/C.
The direction of E_net can be found by taking the arctan:
tanθ = (E_C / E_A) = (5.15 x 10^4) / (8 x 10^4) = 0.644,
θ = arctan(0.644) = 32.2°.
Therefore, the magnitude of the electric intensity at corner B is 9.46 x 10^4 N/C, and the direction is 32.2° below the line AB.