Question

A straight, nonconducting plastic wire 7.00 cmlong carries a charge density of 100 nC/mdistributed 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 4.00 cm directly above its midpoint.

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 4.00 cm directly above its center.

Answer 1

Case a)

This problem can be classified as the obtaining of an electric field due to a charged bar. Where the bar has a length L and a charge density and we will have to calculate the electric field E at a given point. This could be modeled by the next formula:

E = k ( ) ( (1/a) - (1/ L +a))  

Where:

k = 9 x 10 ^9 N m^2 / C ^2 , L = 0.07m , = 100 x 10 ^-9 C / m , and a = 0.005 m because the point is located .04 m directly above its midel point (0.035m) , that is : the point is located 0.05m out of the wire (.04 - .035 = 0.005m ), so substituting this values in the previous formula:

E = k ( ) ( (1/a) - (1/ L +a))   = (9 x 10 ^9 N m^2 / C ^2) ( 100 x 10 ^-9 C / m ) ((1/.005m) - (1/ .07 +.005m))

E = 192 000 N / C

The direction of the Electric Field is along the axis -x to the left.

Case b)

Now the problem is type of electric field of a uniformly charged ring. And we can apply the following formula:

E = k(x) / ((a^2 + x^2)^3/2)) ( Q)   Where:

k = 9 x 10 ^9 N m^2 / C ^2

a = radius of the wire = .035m

x = distance of point P above the ring´s center = 0.04 m

and Q = 100 x10^-9 C

So, substituting values:

E = k(x) / ((a^2 + x^2)^3/2)) ( Q) = ( 9 x 10 ^9 N m^2 / C ^2)(0.04m) / ((.035m)^2+(.04^m)^2) ^3/2 ( 100 x 10 ^-9 C)

E= 239, 758.947 N /C

The direction of the electric field is along the x-axis