1.6C: Una varilla larga y cargada

Una vara larga lleva una carga de\(λ\) coulombs per metre of its length. What is the strength of the electric field at a point P at a distance \(r\) from the rod?

\(\text{FIGURE I.2}\)

Considerar un elemento\(δx\) of the rod at a distance \((r^2 + x^2 )^{1/2}\) from the rod. It bears a charge \(λ\) \(δx\). The contribution to the electric field at P from this element is \(\frac{1}{4\pi\epsilon_0}\cdot \frac{\lambda \delta x}{r^2 + x^2}\) in the direction shown. The radial component of this is \(\frac{1}{4\pi\epsilon_0}\cdot \frac{\lambda \delta x}{r^2 +x^2}\cos θ\). But \(x=r\tan \theta,\, \delta x=r\sec^2 \theta \,\delta \theta \text{ and }r^2+x^2 = r^2\sec^2 \theta\). Therefore the radial component of the field from the element \(δx\) is \(\frac{\lambda}{4\pi\epsilon_0 r}\cos \theta \, \delta \theta\). To find the radial component of the field from the entire rod, we integrate along the length of the rod. If the rod is infinitely long (or if its length is much greater than r), we integrate from \(θ = −π/2 \text{ to }+ π/2\), or, what amounts to the same thing, from \(0 \text{ to }π/2\), and double it. Thus the radial component of the field is

\[\tag{1.6.8} E=\frac{2\lambda}{4\pi\epsilon_0 r}\int_0^{\pi/2}\cos \theta \, \delta \theta = \frac{\lambda}{2\pi\epsilon_0 r}.\]

El componente del campo paralelo a la varilla, por consideraciones de simetría, es cero, por lo que la Ecuación 1.6.8 da el campo total a una distancia\(r\) from the rod, and it is directed radially away from the rod.

Observe que la Ecuación 1.6.4 para una distribución de carga esférica tiene\(4πr^2\) en el denominador, mientras que la Ecuación 1.6.8, que trata de un problema de simetría cilíndrica, tiene\(2πr\).