Question

1. -12.27 points KatzPSEf1 24.P.011 My NotesAsk You Given the two charged particles shown in the figure below, find the electric field at the origin. (Let q1 =-24.00 nC and q2 = 6.00 nC. Express your answer in vector form.) E= N/C y (cm) 91 72 x(cm) Need Help? L Read

Answer 1

Diagram: y(cm) 9 E sin E, sino E E cose ços x(cm) 4 Esind The electric field at the origin due to the charge q = -24.00 nC is, Ē, = (E. cose)+(E, sino)j ........ (1) Here, E = kg, And 2 cm cos = *** H(2 cm) +(3 cm) 3 cm sin = T(2 cm)+ (3 cm)

Therefore, the equation (1) changes as follows: |(9x10° N-m?/C?)(24x10ºC) (2x10-2 m) } (2x102m) +(3x10^2 m)" (2x10^2 m)? +(3x10-2 m)? | [{(9x10° N·m? /C?)(24x10~9c) (3x10^2 m) lil "11 (2x10^2m)” +(3x102 m) (2x10-2 m) +(3x10-2 m)"J). = (9.216x10* N/C) î +(1.38248x10N/C) Î The electric field at the origin due to the charge qz = 6.00 nC is, Ē, = (E, cosø)ỉ +(E, sino)(-1) Here, .... (2) And COSO=Tacm 4 cm (4 cm) + (1 cm) +(1 cm 1 cm sin = (4 cm) + (1 cm)

Therefore, the equation (2) changes as follows: ((9x10° N·m²/C2)(6x10-c) (4x10-2 m) (4x10^2m) + (1x102 m)" (4x102 m)*+(1x10^2 m)° ) [[(9x10° N·mº/C?)(6x10ºC) (1x102 m) ll (4x10^m) +(1x10-2 m) (4x10-2 m)* +(1x102 m)"") = (3.0816x10* N/C) î+(7.704x10' N/C)(-3) Therefore, the net electric field at the origin due to both charges is as follows: Ē =Ē, +Ē, = (9.216x10* N/C)i +(1.38248x10° N/C) +(3.0816x10* N/C){ +(7.704x10° N/C)(-i) =((1.23x10>)i +(1.305x109))N/C