A long straight wire has a uniform linear charge density of 4.0 times 10 to the negative six coulombs per meter. Calculate the magnitude of the electric field strength at a 2.0-meter perpendicular distance from the axis of the wire.
This wire’s uniform charge density of 4.0 times 10 to the negative six coulombs per meter we’ll call 𝜆. And the perpendicular distance from the axis of the wire, 2.0 meters, we’ll call 𝑑. We want to calculate the magnitude of the electric field strength at a distance of 𝑑 from the axis of the wire.
We can call that magnitude capital 𝐸. If we draw a sketch of this charged wire, we can draw it as extending out very far in both directions of it ends since the problem statement says it’s a long straight wire. We’re told to consider a point perpendicularly located from the wire at distance of 𝑑 or 2.0 meters. It’s at that point that we want to solve for the electric field 𝐸.
To get us started, we can recall the mathematical relationship for the electric field that’s created by a long wire. That electric field is equal to the linear charge density, 𝜆, over two 𝜋 times 𝜖 naught, the permittivity of free space, times 𝑟, which is the distance from the wire the field 𝐸 is being calculated.
Notice that this electric field is symmetric around the ends of the wire. If we looked at a charge-carrying wire and on. The electric field created by that wire would point out equally in all directions. The field diminishes with distance 𝑟, but has nothing to do with the direction from the wire in which it points.
Applying this relationship to our scenario, instead of 𝑟 we use distance 𝑑. And for 𝜖 naught, which is a constant, we use an exact value of 8.85 times ten to the negative 12 farads per meters. We’re now ready to plug it in and solve for 𝐸.
When we enter these values on our calculator, we find that, to two significant figures, the electric field magnitude is 3.6 times ten to the fourth newtons per coulomb. That’s how strong the electric field is 2.0 meters from this wire.