Electric Field Due to a Uniformly Charged Arc
Electric Field Due to a Uniformly Charged Arc
Consider a uniformly charged arc of radius \(R\), total charge \(Q\), and angle subtended \(\alpha\) at the center. We derive the electric field at the center of the arc.
Electric Field
Only the horizontal components survive due to symmetry.
\[dE_x = dE\cos\theta\] \[dE_x=\frac{k\,dq}{R^2}\cos\theta\] \[dE_x=\frac{k}{R^2}\,\frac{Q}{R.\alpha}.dl\cos\theta\] \[dE_x=\frac{k}{R^2}\,\frac{Q}{R.\alpha}. R d\theta\cos\theta\] \[E=\frac{k}{R^2}\,\frac{Q}{\alpha} \int_{-\alpha/2}^{+\alpha/2}\cos\theta\, d\theta\] \[E=\frac{k}{R^2}\,\frac{Q}{\alpha} \left[\sin\theta\right]_{-\alpha/2}^{+\alpha/2}\] \[E=\frac{k}{R^2}\,\frac{Q}{\alpha} \left(\sin\frac{\alpha}{2}+\sin\frac{\alpha}{2}\right)\] \[\boxed{E=\frac{2kQ}{R^2\alpha} \sin\frac{\alpha}{2}}\]In terms of $\lambda$ Linear Charge Density
\[\lambda = \frac{Q}{R\,\alpha}\] \[E=\frac{2k\lambda}{R}\sin\frac{\alpha}{2}\]
Electric Potential
All charge elements are at the same distance \(R\)from the center.
\[dV=\frac{k\,dq}{R}\] \[V=\int \frac{k\,dq}{R}\] \[V=\frac{k}{R}\int dq\] \[\boxed{V=\frac{kQ}{R}}\]