Question

What is the strength of the electric field at the position indicated by the dot in (Figure 1)? You may want to review (Pages 846- 648). N/C Figure 1 of 1 1.0 nC 5.0 cm Part B What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line. Express your answer using two significant figures. 5.0 cm 5.0 cm 1.0 nC

Answer 1

The magnitude of the net electric field is 180N. Which is the vectorial sum of the two field at P (or dot) due to both charges. And by seeing the figure using Algebra we can find the direction of the resultant is 90 degree below the horizontal line.

Let p be the position of the dot. electric field at p due to inc is, E^vector_p_1 = 1/4 pi epsilon_0 times q/r^2 e_r^^= 9 times 10^9 times 1 times 10^-9/5 squareroot 2 times 10^-2 E^vector_p_1 = 1.276 times 10^2 N/C along AP electric field at p due to -1 nc is E^vector_p_(-1) = 1/4 pi epsilon_0 times q/r^2 = 9 times 10^9 times 1 times 10^-9/5 squareroot 2 times 10^-2 = 1.276 times 10^2 N/C along PB So net field at P is, now the angle between E^vector_p_1 and E^vector_p_(-1) is pi/2 or 90 degree E^vector_Net = squareroot = (E_p_1)^2 + (E_p_(-1))^2 + 2E_p_1 middot E_p_(-1) cos pi/2 = squareroot (1.276 times 10^2)^2 + (1.276 times 10^2)^2 = squareroot 32563.52 N = 180 N (B) Direction of the field is 90 degree below the Horizontal line. [from the geometry].