A uniformly charged rod extends from y 150 mm to y 150 mm

A uniformly charged rod extends from y = -150 mm to y = +150 mm along the y axis of an xy coordinate system. The charge on the rod is q = 36 nC. With the goal of eventually calculating the electric field magnitude along the rod\'s perpendicular bisector, approximate the rod as three charged particles: two at the ends of the rod and one in the middle, each carrying the charge of q/3. What is the electric field magnitude at x = 200 mm using this approximation? Express your answer with the appropriate units. What is the error in your result in the previous part?

Solution

due to middle charge:

E1 = k (q/3) / x^2

E1 = (9 x 10^9 ) (36 x 10^-9 / 3) / (0.200)^2

E1 = 2700 N/C away from rod.

due to end charge.

magnitude of field will be same due to both charge.

E = k (q/3) / (y^2 + x^2)

E = (9 x 10^9)(36 x 10^-9 / 3) / (0.200^2 + 0.150^2)

E = 1728 N/C

for resultant of these two, component along the length will cancel out and perpendicular to the rod will get added.

E2 = 2 E sin@

where sin@ = x / sqrt ( x^2 + y^2)

= 0.200 / sqrt(0.150^2 + 0.200^2) = 0.8

E2 = 2 x 1728 x 0.8 = 2764.8 N/C

Enet = E1 + E2 = 5464.8 N/C ............Ans

(B) E = k Q / x sqrt(y^2 + x^2)

Enet = (9x 10^9 x 36 x 10^-9) / (0.200 sqrt(0.200^2 + 0.150^2))

= 6480 N/C

deltaE / Eexact = | 6480 - 5464.8 | / 6480 = 0.157