Question

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.

Part A

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

Part 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.

Answer 1

Call the linear charge desnity l, the length of the wire L and the distance above the wire h. Define x to be the direction along the wire. Let x = 0 be at the center of the wire.

Now if you choose a line segment of length dx at a distance +/- x from the center, you can see by symmetry that the electric field at h muct point vertically up (the components of the filed in the x direction cancel). So we can write the vertical component dE as:

dE = l dx/(x^2 + h^2) * cos(Q) where Q = angle between the E field vector and the vertical at h.

But by trig, we can write cos(Q) as

cos(Q) = h/sqrt(x^2 + h^2)

so

dE = l*h*dx/(x^2 + h^2)^3/2 now integrate over x from -L/2 to L/2

E = l*h*x/(h^2*sqrt(x^2+h^2)) [-L/2,L/2] =2*l*L/(h*sqrt((L/2)^2 +h^2))

For a circle of radius a = L/(2*pi), again only the vertical component of the field survives at h, But now the field takes the form:

dE = l*a*dw/(a^2 +h^2) *cos(Q') where cos(Q') = h/sqrt(a^2 + h^2) & dw = angle in plane of circle

so

dE= l*a*h*dw/(a^2+h^2)^3/2

Integrate from 0 to 2*pi

E = 2*pi*l*a*h/(a^2+h^2)^(3/2) = l*L*h/((L/2)^2 +h^2)^3/2