Question

Consider the symmetrically arranged charges in the figure, in which q a = q b = − 1.55 μC and q c = q d = + 1.55 μC . Four charges located at the corners of a square. Charge q subscript a is at the upper left corner. Charge q subscript b is at the upper right corner. Charge q subscript c is at the lower left corner. Charge q subscript d is at the lower right corner. A fifth charge q is located at the exact center of the square. Determine the direction of the electric field at the location of charge q . left down and left down and right up and right down right up and left up Calculate the magnitude of the electric field E at the location of q given that the square is 5.35 cm on a side. E = N / C Question Credit: OpenStax College Physics

Answer 1

5.35cm K=9x109 Nm2 ca Electrostatic = constant charges at A and B = 4 =-1.ssue = -1.55x100c Charges at a and D= q = 1.5sx 1000 e sides of square =AB= AC=BD = CD = 5.35 cm = 0.0935 m. length of diagonal B = d . d = JAC? + cor = r AC = 0.076m . f is at the same distance from A, B, C, D. which is r=d = 0.038 m. Electric field (fot og due to A = E = kq (along AD towards A) E F at a due to D - E = kar (along AD towards A) net EF at a due to A and D E = 2 x 9 x 10°xlossX106 Nic = E + Ez (0.0382 E is towards A along AD. E = 1.93 x 108 Nie Similarly, Ef at a due to B and C is E = 1.93 x 108 N/C (along BC towards B) neet Efat q = Enet = JEZ+E (El=1Ella and they are at 90°) = 2 x (1.93% 108 ) = (1.938 108) Xu Enet = 2.73 × 10S NIC (parallel to AC, ataing towards AB )