A 1.96 10-9 C charge has coordinates x = 0, y = ?2.00; a 2.97 10-9 C charge has

Question

A 1.96 10-9 C charge has coordinates x = 0, y = ?2.00; a 2.97 10-9 C charge has coordinates x = 3.00, y = 0; and a -5.15 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. A. Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis). B. 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

i know answers are wrong i help you i give ou the process change values and calculate and give me good rating The field due to the 1st charge is in the +y direction F1 = k*q1/r1^2 = 9.0x10^9*1.92x10^-9/0.02^2 = 43200N/C (+y) F2 is in the -x direction F2 = k*q2/r2^2 = 9.0x10^9*3.24x10^-9/0.030^2 = 32400N/C (-x) F3 points in the -x & -y directions F3 = k*q3/r3^2 = 9.0x10^9*5.20x10^-9/0.050^2 = 18720N/C For this point cos(?) = -0.60 and sin(?) = -0.80 F3x = 18720*(-0.60) = -11232N/C F3y = 18720*(-0.80) = -14976N/C So Fx = -32400-11232 = -43632N/C Fy = 43200 - 14976 = 28224N/C So F = sqrt(43632^2 + 28224^2) = 5.20x10^4N/C ? = arctan(28224/-43632) = -32.897 but ? = -32.897 + 180 = 147o (since Fx is negative) b) a = F/m = E*q/m = 5.20x10^4*1.60x10^-19/1.67x10^-27 = 4.98x10^12m/s^2 the angle is the same as in a 147o

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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