(a) Determine the magnitude and direction for the electric field at the origin (

Question

(a) Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis).

magnitude direction A 2.20 times 10-9 C charge has coordinates x = 0, y = ?2.00; a 2.91 times 10-9 C charge has coordinates x = 3.00, y = 0; and a -4.70 times 10-9 C charge has coordinates x = 3.00, y = 4.00, where all distances are in cm. Determine magnitude and direction for the electric field at the origin and the instantaneous acceleration of a proton placed at the origin. Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis). Determine the magnitude and direction for the instantaneous acceleration of a proton placed at the origin (measure the angle counterclockwise from the positive x-axis).

Explanation / Answer

The field due to the 1st charge is in the +y direction

F1 = k*q1/r1^2 = 9.0x10^9*2.2x10^-9/0.02^2 = 49500N/C (+y)

F2 is in the -x direction

F2 = k*q2/r2^2 = 9.0x10^9*2.91x10^-9/0.030^2 = 29100N/C (-x)

F3 points in the -x & -y directions

F3 = k*q3/r3^2 = 9.0x10^9*4.70x10^-9/0.05^2 = 16920N/C
For this point cos(?) = -0.60 and sin(?) = -0.80
F3x = 16920*(-0.60) = -10152N/C
F3y = 16920*(-0.80) = -13536N/C

So Fx = -29100-10152 = -39252N/C

Fy = 49500 - 13536 = 35964N/C

So F = sqrt(39252^2 + 35964^2) = 5.320x10^4N/C

? = arctan(35964/(-39252)) = -42.49 but ? = -42.49 + 180 = 137.5 deg (since Fx is negative)


b) a = F/m = E*q/m = 5.32x10^4*1.60x10^-19/1.67x10^-27 = 5.09x10^12m/s^2

the angle is the same as in a 137.5 deg

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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