A 2.06 10-9 C charge has coordinates x = 0, y = ?2.00; a 3.12 10-9 C charge has

Question

A 2.06 10-9 C charge has coordinates x = 0, y = ?2.00; a 3.12 10-9 C charge has coordinates x = 3.00, y = 0; and a -4.95 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

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

F1 = k*q1/r1^2 = 9.0x10^9*2.06x10^-9/0.02^2 = 46350 N/C (+y)

F2 is in the -x direction

F2 = k*q2/r2^2 = 9.0x10^9*3.12x10^-9/0.030^2 = 31200 N/C (-x)

F3 points in the -x & -y directions

F3 = k*q3/r3^2 = 9.0x10^9*4.95x10^-9/0.050^2 = 17820 N/C
For this point cos(?) = -0.60 and sin(?) = -0.80
F3x = 17820*(-0.60) = -10692 N/C
F3y = 17820*(-0.80) = -14256 N/C

So Fx = -31200-10692 = -41892 N/C

Fy = 46359 - 14256 = 32094N/C

So F = sqrt(41892^2 + 32094^2) = 5.27x10^4N/C

? = arctan(32094/-41892) = -37.46 but ? = -37.46 + 180 = 142.54o (since Fx is negative)


b) a = F/m = E*q/m = 5.27x10^4*1.60x10^-19/1.67x10^-27 = 5.05x10^12m/s^2

the angle is the same as in a 142.54o

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