Three point charges of q 1 =-3µC, q 2 =+4 µC, and q 3 =+5mC respectively are loc

Question

Three point charges of q1=-3µC, q2=+4 µC, and q3=+5mC respectively are located as shown on the left at (0.0m, 2.0m), at (2.0m, 0.0m), and at (-1.0m, 0.0m). We would like to know the electric field at point P (2,2).

(a) Determine the electric field vector E1 due to charge q1

    (i.e. you need magnitude and direction!)

(b) Determine the electric field vector E2 due to charge q2

(c) Determine the electric field vector E3 due to charge q3

(d) Determine the total electric field E , i.e. the sum of the electric fields E1 and E2 and E3.

Explanation / Answer

q1 = -3*10^-6 C at (0,2) q2 = 4*10^-6 C at (2,0) q3 = 5*10^-3 C at (-1,0) P(2,2) be the point at which we have to calculate the resultant Electric field. Seperation between q1 and P, r1 = Sqrt[(2-0)^2 + (2-2)^2]                                                    = Sqrt[4 + 0]                                                    = 2 m Seperation between q2 and P, r2 = Sqrt[(2-2)^2 + (2-0)^2]                                                    = Sqrt[0 + 4]                                                    = 2 m Seperation between q2 and P, r2 = Sqrt[(2-2)^2 + (2-0)^2]                                                    = Sqrt[0 + 4]                                                    = 2 m      Seperation between q3 and P, r3 = Sqrt[(2+1)^2 + (2-0)^2]                                                    = Sqrt[9 + 4]                                                    = 3.605 m (a) Electric field due to q1, E1 = (Kq1/r1^2) [ - i ] Seperation between q3 and P, r3 = Sqrt[(2+1)^2 + (2-0)^2]                                                    = Sqrt[9 + 4]                                                    = 3.605 m (a) Electric field due to q1, E1 = (Kq1/r1^2) [ - i ]                                          = (9*10^9 * 3*10^-6/4) [ - i ]                                          = - 6.75*10^3 i N/C (b) Electric field due to q2, E2 = (Kq2/r2^2) [j ]                                          = (9*10^9 * 4*10^-6/4) [ j ]                                          = 9*10^3 j N/C (c) Electricv field due to q3, E3 = (Kq3/r3^2) [cos33.69 i + sin33.69 j]                                            = (9*10^9 * 5*10^-3/13] [0.832 i + 0.554 j]                                            = 3.461*10^6 [0.832 i + 0.554 j]                                            = [2880 i + 1917.1j]*10^3 N/C (d)   Net electric field E = E1 + E2 + E3                              =  - 6.75*10^3 i +  9*10^3 j + [2880 i + 1917.1j]*10^3                              = [2873.25 i + 1926.1 j]*10^3 N/C Magnitude |E| = 3459.1*10^3 N/C Direction = Tan^-1[1926.1/2873.25]                   = 33.830      degrees anticlock wise from the +ve X axis.    Direction = Tan^-1[1926.1/2873.25]                   = 33.830      degrees anticlock wise from the +ve X axis.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.