Template for three columns and two columns

Self Energy of a Uniformly Charged Spherical Shell and Solid Sphere

Three Column and two column template title region

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.

\[dV=\frac{k\,dq}{R}\] \[V=\int \frac{k\,dq}{R}\] \[V=\frac{k}{R}\int dq\] \[\boxed{V=\frac{kQ}{R}}\]

Special Cases

Quarter Circle, $\alpha = \frac{\pi}{2}$ \[E=\frac{2\sqrt{2}kQ}{\pi R^2}\] Semicircle, $\alpha = \pi$ \[E=\frac{2kQ}{\pi R^2}\] Full Circle $\alpha = 2\pi$ \[E = 0\] \[V = \frac{kQ}{R}\]

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.

\[dV=\frac{k\,dq}{R}\] \[V=\int \frac{k\,dq}{R}\] \[V=\frac{k}{R}\int dq\] \[\boxed{V=\frac{kQ}{R}}\]