Question

A)A straight, nonconducting plastic wire 7.50cm long carries a charge density of 130nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.

-Find the magnitude and direction of the electric field this wire produces at a point 5.00cm directly above its midpoint.

B) pick one of the two - electric field is directed upward - electric field is directed downward C) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 5.00cm directly above its center. D) Pick one of the two - electric field is directed upward - electric field is directed downward

Answer 1

PART B:

The magnitude of the electric field at P due to the segment of charge dq is

dE = k dq/r²

This field has an x component dE_x = dE cos? along the axis and a component dE perpendicular to the axis. however, the resultant field at P must lie along the x axis because the perpendicular components of all the various charge segments sum to zero. That is, the perpendicular component of the field created by any charge element is canceled by the perpendicular component created by an element on the opposite side of the ring. Because r=(x²+a²)^1/2 and cos? = x/r, we find that

dEx = dE cos? = dE = (k dq/r²) (x/r) = k x/(x²+a²)^3/2 dq

All segments of the ring make the same contribution to the field at P because they are all equidistant from this point. Thus, we can integrate to obtain the total field at P :

E = ?k x/(x²+a²)^3/2 dq = k x/(x²+a²)^3/2 ?dq = k Q x/(x²+a²)^3/2

Therefore,

Q = 130 nC/m * 7.50e-2 = 9.75 nC

radius ofthe cicle = a = L/2? = 7.50e-2/(2*3.14) = 0.0119427

E = kxQ/(x²+a²)^3/2 = (8.99e9*5.5e-2*9.75e-9)/(5.5e-2*5.5e-2+0.0119427*0.0119427)^3/2 = 27041 N/C