Solid angle is not in the Class 12 NCERT syllabus; JEE Advanced uses it. A Class 12 student needs two results of this lesson. A point charge sends \(q/\varepsilon_0\) through a sphere around it. A charge at the centre of a cube sends a sixth of that through each face. The rest of the lesson shows where those results come from.
An angle in three dimensions
A plane angle is an arc divided by its radius, \(\theta = l/r\), in radians. A full circle is \(2\pi r/r = 2\pi\) radians. The size of the circle does not matter, because a larger circle has a longer arc in the same ratio. A cone that opens from a point needs the same idea in three dimensions.
Draw a sphere of radius \(r\) around the tip of a cone. The cone cuts an area \(A\) out of the sphere. The solid angle of the cone is \(\Omega = A/r^2\), measured in steradians (sr). A sphere twice as large has four times the area inside the same cone, so the ratio stays fixed. The whole sphere has \(\Omega\) \(= 4\pi r^2/r^2\) \(= 4\pi\,\text{sr}\). The half of the sky above a flat field is \(2\pi\,\text{sr}\).
A small patch that does not face the tip squarely looks smaller from the tip. Let its normal make an angle \(\theta\) with the line from the tip. Its shadow across that line is \(dA\cos\theta\), so it holds the solid angle \(d\Omega = dA\cos\theta/r^2\). Every patch with the same edge seen from the tip holds the same solid angle, whatever its shape.
Open the cone with the slider. At a half-angle of \(60^\circ\), \(\Omega\) \(= 2\pi(1 - \tfrac12)\) \(= \pi\,\text{sr}\), a quarter of the sphere. At \(90^\circ\) the cone has become a flat plane, and \(\Omega = 2\pi\), half the sphere. At \(180^\circ\) it closes round the whole sphere, and \(\Omega = 4\pi\).
A flat patch of area \(1\,\text{cm}^2\) faces a point squarely from a distance of \(10\,\text{cm}\). What solid angle does it hold at the point, in steradians?
The flux of a point charge
In the chapter Electric Charge, Coulomb's Law and the Electric Field, a point charge \(q\) makes a field \(E = q/4\pi\varepsilon_0 r^2\), pointing straight away from it. Surround the charge by a sphere of radius \(r\). On the sphere the field has one size everywhere and crosses along the outward normal. So the flux is the field times the whole area.
\[\Phi = \frac{q}{4\pi\varepsilon_0 r^2} \times 4\pi r^2 = \frac{q}{\varepsilon_0}\] The radius has cancelled. A sphere twice as large has four times the area, and the field on it is four times weaker. The same field lines cross every sphere around the charge, so the count of lines cannot depend on the sphere.
A charge of \(53.1\,\text{nC}\) sits at the centre of two spheres, of radii \(5\,\text{cm}\) and \(10\,\text{cm}\). Find the field on each sphere and the flux through each. Take the \(\varepsilon_0\) route.
On the small sphere, \(E\) \(= q/4\pi\varepsilon_0 r^2\) \(= 53.1 \times 10^{-9}/(4\pi \times 8.85 \times 10^{-12} \times 0.0025)\) \(\approx 1.91 \times 10^5\,\text{N/C}\). Its area is \(4\pi(0.05)^2 = 0.0314\,\text{m}^2\), so \(\Phi = 6000\,\text{N m}^2/\text{C}\).
On the large sphere the field is four times weaker, \(4.8 \times 10^4\,\text{N/C}\), and the area four times larger. The flux is again \(6000\,\text{N m}^2/\text{C}\).
Both equal \(q/\varepsilon_0\) \(= 53.1 \times 10^{-9}/8.85 \times 10^{-12}\) \(= 6000\,\text{N m}^2/\text{C}\). This charge was chosen to make that number round.
Values used: \(\varepsilon_0\) \(= 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)\).
Two spheres of radii \(r\) and \(2r\) are drawn around the same point charge. What are the ratios, small sphere to large, of the flux through them and of the field on them?
Sharing the flux by solid angle
The field of a point charge is the same in every direction. So its flux \(q/\varepsilon_0\) is spread evenly over the \(4\pi\) steradians around it. A cone of solid angle \(\Omega\) with its tip at the charge takes the share \(\Omega/4\pi\). Any surface that fills that cone, flat or curved, takes the same flux, because every line in the cone crosses it once.
For the charge with \(q/\varepsilon_0\) \(= 6000\,\text{N m}^2/\text{C}\), a cone of half-angle \(60^\circ\) holds \(\pi\) sr, a quarter of the sphere. So \(1500\,\text{N m}^2/\text{C}\) passes through it. The share is pure geometry. No value of \(k\) or \(\varepsilon_0\) enters it, only the shape of the cone.
The charge with \(q/\varepsilon_0\) \(= 6000\,\text{N m}^2/\text{C}\) sits on the axis of a disc of radius \(3\,\text{cm}\), \(4\,\text{cm}\) from its centre, as in Fig 2.1. Find the flux through the disc.
The disc fills a cone with its tip at the charge. The half-angle \(\theta\) runs from the axis to a line to the rim.
\(\cos\theta = 4/5 = 0.8\), and \(\Omega = 2\pi(1 - \cos\theta)\).
\(\Omega\) \(= 2\pi(1 - 0.8)\) \(= 0.4\pi\) \(\approx 1.26\,\text{sr}\). The share is \(0.4\pi/4\pi = 0.1\), so \(\Phi\) \(= 0.1 \times 6000\) \(= 600\,\text{N m}^2/\text{C}\).
A tenth of the charge's flux passes through the disc. Moved closer to the charge, the disc would hold a wider cone and take more.
The same charge, with \(q/\varepsilon_0\) \(= 6000\,\text{N m}^2/\text{C}\), sits on the axis of a disc of radius \(4\,\text{cm}\), \(3\,\text{cm}\) from its centre. What is the flux through the disc, in N m²/C?
A charge in a cube
Put the charge at the centre of a cube. Every line from the charge leaves through one face, so the whole cube carries \(q/\varepsilon_0\) \(= 6000\,\text{N m}^2/\text{C}\). The six faces look exactly alike from the centre. So each face takes a sixth, \(1000\,\text{N m}^2/\text{C}\). This answer needs symmetry, not an integral.
Now move the charge to a corner. Eight cubes meet at that corner and share the space around it equally, so each takes an eighth: \(6000/8\) \(= 750\,\text{N m}^2/\text{C}\). The three faces that touch the charge take nothing, because the field runs along them. The three far faces are alike, so each takes \(750/3 = 250\,\text{N m}^2/\text{C}\).
The same trick handles a single square. Put the charge at a height \(a/2\) above the centre of a square of side \(a\). Build a cube of side \(a\) on the square. The charge is at its centre, so the square takes a sixth, \(1000\,\text{N m}^2/\text{C}\). The imaginary faces change nothing, because they carry no charge of their own.
A charge \(q\) sits at one corner of a cube. What is the flux through one face of the cube that does not touch the charge?
The charge with \(q/\varepsilon_0\) \(= 6000\,\text{N m}^2/\text{C}\) sits at the centre of one face of a cube. What is the total flux through the cube, in N m²/C?
The box of an LED bulb gives its light output in lumens, say \(800\,\text{lm}\). The brightness in one direction is given in candela, and one candela is one lumen per steradian. A bulb that shone equally in all directions would give \(800/4\pi \approx 64\,\text{cd}\). A reading lamp with a shade sends the same light into a smaller solid angle, so it looks brighter below the shade.
A bare bulb hangs \(1\,\text{m}\) from a book whose cover of \(100\,\text{cm}^2\) faces it. Roughly what percentage of the bulb's light falls on the cover?
- Using the half-angle of a cone as its solid angle.The solid angle is \(2\pi(1 - \cos\theta)\) steradians; a half-angle of \(90^\circ\) holds \(2\pi\), not \(\pi/2\).
- Giving each face of a cube a sixth of the flux when the charge is not at the centre.A sixth per face needs the charge at the centre. Elsewhere, complete the space with more cubes and use symmetry.
- Thinking that a larger sphere around a charge carries more flux.The field falls as \(1/r^2\) while the area grows as \(r^2\), so every sphere carries \(q/\varepsilon_0\).