IA For the charges in the previous problem, if the negative charge sits at x = 0
ID: 1771847 • Letter: I
Question
IA For the charges in the previous problem, if the negative charge sits at x = 0 and the cm, determine the place(s) on the x-axis where the net electric field due to these two charges is zero. (x +0.337 ositive charge sits at x-2.5 m = 33.7 cm) Four electric charges sit at the corners of a square 51.4 cm on a side. One charge is +40.0-uc and the other three are -23.0-uC charges. Determine the magnitude of the net electric field at the center of the square. (4.29 x 10 N/C, directly away from positive charge] " A long thin rod of length 13.0 m lies along the x-axis with its center at the origin. The total charge on the rod is -120 pC Determine the net electric field at a point 30.0 m above the rod on the y-axis. (Hint: This is an integral problem. The solution is not simply E = .) [1.17 x 10, N/C] ·P 30.0 mExplanation / Answer
B)The two digonally apposite charges will not contribute to electric field as the ellectric field at the center will cancel because they are same magnitude and distance of the charges.
The field at the center due to +40 is
r = sqrt (2*51.4^2)/2 = 36.35 cm = 0.364 m
E1 = E1 cos(theta) i + E1 sin(theta) j
E1 = 9 x 10^9 x 40 x 10^-6 x cos45/0.364^2 + 9 x 10^9 x 23 x 10^-6 x sin45/0.364^2
E1 = 1.92 x 10^6 i + 1.92 x 10^6 j
E2 = E3 cos(theta) + E3 sin(theta)
E2 = 9 x 10^9 x 23 x 10^-6 x cos45/0.364^2 + 9 x 10^9 x 23 x 10^-6 x sin/0.364^2
E2 = 1.11 x 10^6 i + 1.11 x 10^6 j
Ex = E1x + E2x = 3. 03x 10^6 i
Ey = Ey1 + Ey2 = 3.03 x 10^6 j
E = sqrt (3.03^2 + 3.03^2) x 10^6 = 4.29 x 10^6 N/C
Hence, E = 4.29 x 10^6 N/C