Question

Two small charged objects are located as shown in the diagram. In the diagram each square represents le-2 m. Q1 =-2 nC and Q2 electric field at location A = {-0.04,-0.04,。> m. 5 nC. Remember that l nC nanocoulomb) is 1e-9 coulombs. You will calculate the net Before doing any calculations, on a copy of the diagram, draw two arrows representing the electric field at the observation location due to each of the charged particles. Use your diagram to answer the following questions What is the sign of the x-component of , the electric field at A de to Q1positive What is the sign of the y-component of E1? positive What is the sign of the z-component of E1? zero What is the sign of the x-component of Bz, the electric field at 7 due to Qz? negative What is the sign of the y-component of Ez? negative What is the sign of the z-component of Eq? 「zero- Now calculate these quantities: What is the electric field at A due to Q1? 61 |-804.98447 | NC

Answer 1

Coordinate of q_1 = > (-0.02, 0.05) m coordinate of q_2 = > (0.03, -0.03.) m coordinate of A^vector = > (-0.04, -0.04) m Electric field due to q_1 = k q_1/r^2_1 r^^_1 r^vector_1 = -0.04 - (0.02) i^^+ (-0.04 - 20.05) r^^r^vector_1 = -0.02 i^^-0.09 j^^ Vertical-bar r^vector_1 Vertical-bar = squareroot 0.02 i^^-0.09 j^^ Vertical-bar r^vector_1 Vertical-bar = squareroot 0.02^2 + 0.09^2 = 0.092195 r^2_1 = 0.0085 m^2, E_1 = 9 times 10^9 times (-2) (10^9)/0.0085 (-0.02/0.092195 i^^-0.08/0.092195 r^^) = 459 i^^+ 2067.2 j^^ Vertical-bar E^vector_1 Vertical-bar = 2117.56 N/C r^vector_2 = (-0.04 - 0.03) i^^+ (-0.04 + 0.03) j^^= -0.07 i^^-0.01 j^^ Vertical-bar r^vector_2 Vertical-bar = 0.0707 r^2_2 = 0.005 E_2 = 9 times 10^9 times 5 times 10^-9/0.005 (-0.07 i^^/0.0707 - 0.01 j^^/0.0707) = -8909.5 i^^-1272.79 j^^ Vertical-bar E^vector_2 Vertical-bar = 9000 N/C E_net = (459 - 8909.5) i^^+ (2067.2 - 1272.79) j^^= -8450 i^^+ 794.41 j^^ Vertical-bar E^vector_net Vertical-bar = 8487.758 N/C