Question

A 2.10 10-9 C charge has coordinates x = 0, y = −2.00; a 3.24 10-9 C charge has coordinates x = 3.00, y = 0; and a -4.70 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).

Magnitude? in N/C

Direction? in degrees

(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).

Magnitude? in N/C

Direction? in degrees

Answer 1

a)
let

q1 = 2.1*10^-9 C
q2 = 3.24*10^-9 C
q3 = -4.7*10^-9 C

Enetx = E1x - E2x + E3x

= 0 - E2 + E3*cos(theta)

= 0 - k*q2/d2^2 + (k*q3/d3^3)*(3/5)

= 0 - 9*10^9*3.24*10^-9/0.03^2 + 9*10^9*4.7*10^-9/(0.03^2+0.04^2)*(3/5)

= -2.22*10^4 N/c

Enety = E1y + E2y + E3y

= k*q1/d1^2 - 0 + E3*sin(theta)

= k*q1/d1^2 + (k*q3/d3^3)*(4/5)

= 9*10^9*2.1*10^-9/0.02^2 + 9*10^9*4.7*10^-9/(0.03^2+0.04^2)*(4/5)

= 6.08*10^4 N/c

Enet = sqrt(Enetx^2 + Enety^2)

= sqrt(2.22^2 + 6.08^2)*10^4

= 6.47*10^4 N/c

direction : theta = 180 - tan^-1(Enety/Enetx)

= 180 - tan^-1(6.08/2.22)

= 110 degrees

b) a = F/m

= q*E/m

= 1.6*10^-19*6.47*10^4/(1.67*10^-27)

= 6.20*10^12 m/s^2

direction : 110 degrees