Option 1 (Bohr's model):
Idea 1: In a hydrogen atom the single electron moves in special circular paths around the nucleus. In each such path the Coulomb attraction supplies the centripetal force, \(\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2} = \dfrac{mv^2}{r}\), and the electron loses no energy while it stays there. These are the stationary, non-radiating orbits.
Idea 2: Not every orbit is permitted. Bohr kept only those for which the electron's angular momentum is quantised, \(mvr = \dfrac{nh}{2\pi}\), with \(n\) a positive integer.
Idea 3: Light is emitted or absorbed only during a jump between two permitted orbits, the photon carrying the energy difference, \(h\nu = E_i - E_f\).
How the drawbacks vanish: Rutherford's accelerating electron would radiate, lose energy and crash into the nucleus in about \(10^{-8}\) s, and would emit every frequency (a continuous spectrum). Because Bohr's electron radiates nothing while in a stationary orbit, the atom stays stable; because it can release only fixed energy differences, hydrogen shows sharp spectral lines instead of a continuous band.
\[\boxed{mvr = \frac{nh}{2\pi},\quad h\nu = E_i - E_f}\]
Option 2 (Field outside a charged hollow sphere):
Step 1: Let a thin spherical shell of radius \(R\) carry charge \(q\) spread evenly over its surface. Choose an imaginary Gaussian sphere of radius \(r\) (with \(r > R\)), concentric with the shell, passing through the external point where the field is wanted.
Step 2: Because the charge distribution is spherically symmetric, the field must point radially and have one and the same magnitude \(E\) everywhere on this Gaussian surface.
Step 3: The electric flux through the Gaussian sphere is \(\Phi = E \cdot (4\pi r^2)\), since \(\vec E\) is everywhere perpendicular to the surface.
Step 4: Gauss's law sets this flux equal to the enclosed charge over \(\varepsilon_0\). The full charge \(q\) lies inside, so \(E \cdot 4\pi r^2 = \dfrac{q}{\varepsilon_0}\).
Step 5: Solving, \(E = \dfrac{q}{4\pi\varepsilon_0 r^2}\). This is exactly the field of a point charge \(q\) at the centre, so an external observer cannot tell a uniformly charged shell from a point charge.
\[\boxed{E = \frac{q}{4\pi\varepsilon_0 r^2}}\]