The figure below shows an uneven arrangement of electrons (e) and protons (p) on
ID: 252058 • Letter: T
Question
The figure below shows an uneven arrangement of electrons (e) and protons (p) on a circular arc of radius r = 1.90 cm, with angles 1 = 30.0°, 2 = 47.0°, 3 = 33.0°, and 4 = 20.0°.
(a) What is the magnitude of the net electric field produced at the center of the arc?
_____ ( degrees / N / C / N/C ) Choose one
(b) What is the direction (relative to the positive direction of the x axis) of the net electric field produced at the center of the arc? (Enter the angle 180° < 180° as measured from the +x axis.)
= ______°
Explanation / Answer
Solution
The field of each charge has magnitude
E = kq/r^2 = k*e/(0.019m)^2
= (8.99* 10^9 N .m^2/ C^2 )*1.60*10^-19C/(0.019m)^2
= 3.98*10^-6 N/C = E
The directions are indicated in standard format below. We use the magnitude-angle notation (convenient if one is using a vector-capable calculator in polar mode) and write (starting with the proton on the left and moving around counterclockwise) the contributions to Enet as follows:
E0+E1+E2+E3+E4 = (Eang0deg)+(Eang210deg)+(Eang260deg)+(Eang130deg)+(Eang340deg)
In component form, it reads as
smesnEx = E0x+ E1x+ E2x+ E3x+ E4x
E cos(0deg ) +cos(210deg )+ cos(260deg )+ cos(130deg )+ cos(340deg )= E 0.257
smesnEy = E0y+ E1y+ E2y+ E3y+ E4y
= E sin(0deg ) sin(210deg ) sin(260deg ) sin(130deg ) sin(340deg ) = E 1.06
a. The result above shows that the magnitude of the net electric field is
|Enet| = sqrt(Ex^2+Ey^2)=sqrt(E^2(0.257)^2+E^2(-1.06)^2) = 3.92*10^8N/C=|Enet|
b. Similarly, the direction of Enet is –76.4deg from the x axis.
theta = taninv (Ex^2+Ey^2) = taninv(E(-1.06)/E(0.257)) = -76.4deg =theta
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