Two charges, Q1 = 2.00 C and Q2 = 6.00 C, are located at points (0,-3.10 cm) and

Question

Two charges, Q1 = 2.00 C and Q2 = 6.00 C, are located at points (0,-3.10 cm) and (0, +3.10 cm), as shown in the figure (For reference, see example 19.4 of the 5th edition of Serway and Jewett's "Principles of Physics" ) 2, what is the magnitude of the electric field at point located at (6.25 cm. 0), due to Q1 alone? Submit Answer Tries 0/10 What is the x-component of the total electric field at P? Submit Answer Tries 0/10 What is the y-component of the total electric field at P? Submit Answer Tries 0/10 What is the magnitude of the total electric field at P? Submit Answer Tries 0/10

Explanation / Answer

distance from charges to point P = r = sqrt(3.1^2+6.25^2) = 6.98 cm = 0.0698 m

magnitude of electric field due to charge Q1, E1 = k*Q1/r^2

E1 = 9*10^9*2*10^-6/0.0698^2 = 3694550.95 N/C <<<-------ANSWER

----------------------------------

magnitude of electric field due to charge Q2, E2 = k*Q2/r^2

E2 = 9*10^9*6*10^-6/0.0698^2 = 11083652.84 N/C

x component of electric due to E1

E1x = E1*x/r = 3694550.95*6.25/6.98 = 3308158.08 N/C

x component of electric due to E2

E2x = E2*x/r = 11083652.84*6.25/6.98 = 9924474.25 N/C

x component of the total elctric field at P Ex = E1x + E2x = 13232632.33 N/C <<<-------ANSWER

_____________________________________________________________________

y component of electric due to E1

E1y = E1*y/r = 3694550.95*3.1/6.98 = 1640846.41 N/C

y component of electric due to E2

E2y = E2*y/r = -11083652.84*3.1/6.98 = -4922539.23 N/C

y component of the total elctric field at P, Ey = E1y + E2y = -3281692.82 N/C <<<-------ANSWER

==========================================

magnitude of electric field at P = sqrt(Ex^2+Ey^2) = 13633490.61 N/C <<<<<-------ANSWER

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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