Show detailed solution k = 9 Times 10^9 Nm^2/C^2, e = plusminus 1.602 Times 10^-
ID: 1425414 • Letter: S
Question
Show detailed solution k = 9 Times 10^9 Nm^2/C^2, e = plusminus 1.602 Times 10^-19 C, F = qE ; E = kq/r^2, F = q_1q_2/r^2 2) Two charges Q_1 and Q_2 are placed 10 cm apart along the axis. Q_1 is at the origin and Q2 is at +10 cm as shown in the diagram above. Q1 = 5 muC and Q2 = -5 muC. At the point P, 12 cm along the Y-axis and midway between the two charges; a) Draw a vector diagram showing the Electric Fields at P due to Q1 and Q2 b) Calculate the magnitude and direction of the electric field ut P due to Q1 c) Calculate the magnitude and direction of the electric field at P due to Q2 d) Combine the two fields and determine the magnitude of the total field at P e) Calculate the direction of the the total field at PExplanation / Answer
Electric field at P due to Q1 = 8.98 * 109 * 5 * 10-6/0.132
= 2656.8 * 103 N/C
Direction = tan-1(12/5)
= 67.38 degrees anticlockwise from + ve x axis .
Electric field at P due to Q2 = 8.98 * 109 * 5 * 10-6/0.132
= 2656.8 * 103 N/C
Direction = - tan-1(12/5)
= - 67.38 degrees clockwise from + ve x axis .
Net electric field at P along x direction = 2 * 2656.8 * 103 * cos67.38
= 2043.7 * 103 N/C
Net electric field at P along y direction = 0 N/C
=> Net electric field at P = 2043.7 * 103 N/C
=> direction of Electric field = towards + ve X axis .