Finding Electric field magnitude of a wire

[Oomair]

Homework Statement

A plastic wire 8.50cm long carries a charge density of +175 nC/m distributed uniformly along its length. it is lying on a horizontal tabletop

A) find the magnitude and direction of the electric field this wire produces at a point 6.00 cm directly above its midpoint

Homework Equations

dq= lamdadx dE= kdq/r^2

The Attempt at a Solution

The part where i am stuck is what should the limits of integration be? 0.00 cm to 6.00 cm?

and what should r^2 be? 6.00 cm?

 

[Defennder]

Note that both r and x are variables; neither are constants. Express one of them in terms of the other before performing the integration. Use pythagoras theorem here. As for limits of integration, it depends on what you choose to vary, either x or r.

 

[baba89]

The limits of integration should be from -4.25cm to +4.25cm, as this represents the distance from the midpoint of the wire to the point 6.00cm above it. The value of r^2 should be (6.00cm)^2, as this represents the distance between the charge element and the point where we are finding the electric field. Therefore, the equation for finding the electric field at this point would be dE = kdq/r^2 * cosθ, where θ is the angle between the charge element and the line connecting it to the point where we are finding the electric field. Since the wire is lying horizontally, θ can be assumed to be 90 degrees, making cosθ = 0. Plugging in the values, the equation becomes dE = (9*10^9 Nm^2/C^2)(175*10^-9 C/m) / (6.00*10^-2 m)^2 * 0 = 0.54 N/C. Therefore, the magnitude of the electric field produced by the wire at the point 6.00cm directly above its midpoint is 0.54 N/C, directed downwards (since the positive charge on the wire will produce an electric field that points away from it).