DEV Community: Surendra Singh Kamboj

Where does the energy go when two capacitors share charge? A JavaScript check

Surendra Singh Kamboj — Sat, 14 Nov 2026 11:59:18 +0000

When a charged capacitor is connected to an empty one, the charge spreads out, but one third of the energy disappears in a typical case. A short simulation shows that the loss does not depend on the resistance of the wire. (I build PractiSci, which has a free capacitor charge-sharing lab.)

What happens when two capacitors share charge?

Take C₁ = 4 µF charged to 12 V and C₂ = 2 µF empty. Charge is conserved: Q = C₁V. After joining, both have the same voltage V′ = C₁V/(C₁ + C₂) = 8 V. Energy is not conserved:

const C1 = 4e-6, C2 = 2e-6, V = 12;
const Ui = 0.5 * C1 * V * V;                 // 288 microjoule
const Vf = C1 * V / (C1 + C2);               // 8 V
const Uf = 0.5 * (C1 + C2) * Vf * Vf;        // 192 microjoule
console.log(Ui - Uf, ((Ui - Uf) / Ui * 100).toFixed(1) + '%');  // 96 microjoule, 33.3%

The loss is 96 µJ, which is 33.3% of the starting energy. It equals ½ · C₁C₂/(C₁ + C₂) · V² = 96 µJ.

Where does the energy go?

It becomes heat in the connecting wire (and some radiation). I check this by integrating I²R over time for two very different resistances:

for (const R of [1, 1000]) {
  const Ce = C1 * C2 / (C1 + C2), tau = R * Ce, dt = tau / 2000;
  let heat = 0, v = V;
  for (let t = 0; t < 12 * tau; t += dt) {
    const i = v / R;
    heat += i * i * R * dt;
    v -= i * dt / Ce;
  }
  console.log('R', R, 'heat', heat.toExponential(4));
}

Both runs give 9.6024 × 10⁻⁵ J, which is the same 96 µJ. A smaller R makes the current larger and shorter, a larger R makes it smaller and longer, but the total heat is identical. Even a wire with almost no resistance loses the same energy (as radiation and sparks).

FAQ

Is charge conserved?

Yes, always. The total charge stays 48 µC. Only the energy is lost.

When is no energy lost?

When the two capacitors are already at the same voltage before they are joined.

Where can I try it?

The free capacitor charge sharing lab lets you change the capacitances and voltage and see the energy before and after. The capacitance guide covers the theory.

Hydrogen spectrum lines from Bohr's model: Balmer wavelengths in JavaScript

Surendra Singh Kamboj — Sat, 14 Nov 2026 11:54:11 +0000

Bohr's model of hydrogen gives every spectral line of the atom from one formula, and you can reproduce the visible Balmer lines in a few lines of JavaScript. Below I compute the wavelengths of the lines you see in a hydrogen lamp and check the photon energy. (I build PractiSci, which has a free Bohr hydrogen lab.)

What is the formula for the hydrogen spectrum?

1/λ = R (1/n_f² − 1/n_i²), where R = 1.0968 × 10⁷ m⁻¹. Electrons falling to n = 2 give the visible Balmer series. Energy levels are Eₙ = −13.6/n² eV.

const R = 1.0968e7;
const nm = (nf, ni) => 1e9 / (R * (1 / nf ** 2 - 1 / ni ** 2));

for (const ni of [3, 4, 5, 6]) console.log(ni, '->2', nm(2, ni).toFixed(1), 'nm');

What wavelengths come out?

Transition	Wavelength	Colour
3 → 2 (Hα)	656.5 nm	red
4 → 2 (Hβ)	486.3 nm	blue-green
5 → 2 (Hγ)	434.2 nm	violet
6 → 2 (Hδ)	410.3 nm	violet

These match the measured Balmer lines. The same function gives 121.6 nm for the Lyman 2 → 1 line (ultraviolet) and 1876 nm for the Paschen 4 → 3 line (infrared), so only the Balmer series is visible.

How do I check it with photon energy?

The 3 → 2 jump releases E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV. The wavelength is λ = 1240 / E nm = 1240 / 1.89 ≈ 656 nm, the same red line.

FAQ

Why is the Balmer series the only visible one?

Its lines end on n = 2 and have energies of about 1.9 to 3.0 eV, which fall in the visible range. The Lyman lines are more energetic (ultraviolet) and the Paschen lines less (infrared).

What is the ionisation energy of hydrogen?

13.6 eV, the energy to take the electron from n = 1 to infinity.

Where can I try it?

In the free Bohr hydrogen lab you excite the atom and watch the lines appear. More Class 12 physics labs are in the labs list.

Orbital speed, escape speed and the square root of 2: a satellite calculator in JavaScript

Surendra Singh Kamboj — Sat, 14 Nov 2026 11:53:18 +0000

The speed a satellite needs to stay in orbit and the speed needed to escape Earth are linked by one number, √2. You can check it in a few lines of JavaScript, and the same code gives the period of the ISS and of a geostationary satellite. (I build PractiSci, which has a free satellite-orbits lab.)

What are the orbital and escape speed formulas?

Orbital speed at height h: v = √(GM/(R + h)). Escape speed: v_esc = √(2GM/(R + h)). Orbital period: T = 2π√((R + h)³/GM).

const G = 6.674e-11, Me = 5.972e24, Re = 6.371e6;
const orbitV = (h) => Math.sqrt(G * Me / (Re + h));
const period = (h) => 2 * Math.PI * Math.sqrt((Re + h) ** 3 / (G * Me));
const escape = (h) => Math.sqrt(2 * G * Me / (Re + h));

What numbers do I get for Earth?

Orbit	Orbital speed	Period	Escape speed
Near the surface	7.91 km/s	1.41 h	11.19 km/s
ISS (400 km)	7.67 km/s	1.54 h	10.85 km/s
Geostationary (35,786 km)	3.07 km/s	23.93 h	4.35 km/s

The ratio escape/orbital is 1.4142 at every height, which is √2. That is why a satellite already has about 70% of the speed it would need to leave Earth altogether.

Why is a geostationary orbit so high?

Its period must equal one sidereal day, 23.93 h. Solving T = 2π√(r³/GM) for r gives r ≈ 42,164 km from the centre of the Earth, which is 35,786 km above the surface. Then the satellite stays over the same point on the equator.

FAQ

Does the mass of the satellite matter?

No. It cancels, so a tiny satellite and a space station at the same height have the same speed.

Why do astronauts feel weightless in orbit?

They are in free fall around the Earth, not beyond gravity. Gravity at 400 km is still about 89% of its surface value.

Where can I try it?

The free satellite orbits lab lets you change the height and see the speed and period, and the work, energy and power guide covers the energy side.

Weak acid titration curve in JavaScript: pH at every point from 0 to 30 mL

Surendra Singh Kamboj — Sat, 14 Nov 2026 11:47:14 +0000

A weak acid titration curve has a buffer region, a half-equivalence point where pH = pKa, and a steep jump at the end point. You can generate the whole curve with four formulas. Below is a JavaScript function for 25 mL of 0.10 M acetic acid titrated with 0.10 M NaOH. (I build PractiSci, which has free virtual titration labs.)

What are the four regions of the curve?

Start: only the weak acid, so [H⁺] = √(Ka·C).
Before the end point: a buffer, so pH = pKa + log([A⁻]/[HA]).
At the end point: only the salt, which hydrolyses, so pH is above 7.
After the end point: excess NaOH sets the pH.

How do I code the curve?

const Ka = 1.8e-5, Kw = 1e-14, Ca = 0.10, Va = 25, Cb = 0.10;

function pH(vb) {
  const na = Ca * Va / 1000, nb = Cb * vb / 1000, V = (Va + vb) / 1000;
  if (vb === 0) {
    const x = (-Ka + Math.sqrt(Ka * Ka + 4 * Ka * Ca)) / 2;
    return -Math.log10(x);
  }
  if (nb < na) return -Math.log10(Ka) + Math.log10(nb / (na - nb));
  if (Math.abs(nb - na) < 1e-12) {
    const c = na / V;
    return 14 + Math.log10(Math.sqrt(Kw / Ka * c));
  }
  return 14 + Math.log10((nb - na) / V);
}

What pH values does it give?

NaOH added (mL)	pH
0	2.88
5	4.14
12.5 (half-equivalence)	4.74
20	5.35
24	6.12
25 (end point)	8.72
26	11.29
30	11.96

At 12.5 mL the pH equals pKa = 4.74, which is how you find Ka from a graph. The end point is at 8.72, not 7, so phenolphthalein (pH range 8.2 to 10) is the right indicator and methyl orange is not.

FAQ

Why is the end point above pH 7?

The acetate ion is a weak base, so the salt solution is slightly basic.

How do I find the end point from the curve?

It is where the curve is steepest, the middle of the vertical jump.

Where can I try it?

The free weak acid titration lab lets you run the titration drop by drop, and the Class 12 chemistry practical list has the other titrations.

Simulating projectile motion with air drag in 25 lines of JavaScript

Surendra Singh Kamboj — Fri, 13 Nov 2026 14:44:39 +0000

Every physics textbook has the projectile formulas for a vacuum: range R = u² sin 2θ / g, time of flight T = 2u sin θ / g and maximum height H = u² sin² θ / 2g. Real balls fly in air, and the answer changes a lot. Here is a small simulation that shows how much, and why the best launch angle is no longer 45 degrees.

The model

Gravity pulls down with g = 9.81 m/s². Air drag points against the velocity and grows with speed. I use a quadratic drag, a_drag = -k * |v| * v, where k (in 1/m) bundles air density, area, mass and the drag coefficient. k = 0 is the textbook case.

The code

function fly({ speed, degrees, k = 0, g = 9.81, dt = 0.001 }) {
  const a = (degrees * Math.PI) / 180;
  let x = 0, y = 0, vx = speed * Math.cos(a), vy = speed * Math.sin(a);
  let t = 0, top = 0;
  while (y >= 0) {
    const v = Math.hypot(vx, vy);
    vx += -k * v * vx * dt;
    vy += (-g - k * v * vy) * dt;
    x += vx * dt;
    y += vy * dt;
    t += dt;
    top = Math.max(top, y);
  }
  return { range: x, time: t, height: top };
}

It is a simple semi-implicit Euler step. With dt = 0.001 s it matches the closed-form answers to about 0.1 percent when k = 0.

What the numbers say

Launch at 20 m/s and 45 degrees:

k (1/m)	range	time of flight	max height
0 (vacuum)	40.8 m	2.88 s	10.2 m
0.02	25.8 m	2.51 s	7.8 m
0.05	17.5 m	2.20 s	6.0 m

The vacuum row agrees with the formulas: 20² x sin 90° / 9.81 = 40.8 m and 2 x 20 x sin 45° / 9.81 = 2.88 s. With a modest drag the range drops by more than a third.

The best angle moves

With k = 0.02 and the same 20 m/s launch, the range is 24.2 m at 30 degrees, 25.4 m at 35 degrees, 25.9 m at 40 degrees and 25.8 m at 45 degrees. The best angle is now a little below 45 degrees, because a flatter shot spends less time in the air losing speed. The more drag, the lower the best angle.

Three things worth remembering for exams

The 45 degree rule holds only for level ground and no air resistance.
Angles theta and 90 degrees minus theta give the same range only in a vacuum.
The horizontal velocity stays constant only without drag.

Play with it

If you would rather drag sliders than edit code, the projectile lab on PractiSci lets you change the speed and angle (with an optional air drag) and plots the path. The chapter guide for projectile motion and kinematics collects the formulas and worked examples for Class 11.

What I learned building 400+ 3D physics and chemistry labs for the browser

Surendra Singh Kamboj — Fri, 13 Nov 2026 14:34:48 +0000

I am building PractiSci, a virtual lab for Class 11 and 12 physics and chemistry students in India. It now has 400+ interactive 3D experiments that run in a normal browser tab, on a phone or a laptop, in English and Hindi. This is what I learned along the way.

The stack

The website is Next.js with three.js for the 3D scenes. The API is Express with MongoDB and JWT sessions. Every lab is a small module that exposes one function which builds the scene, a list of controls and a list of readouts. The player is generic, so a new experiment is mostly a new folder.

Lesson 1: realism is a set of small helpers, not big models

A metre bridge that looks like a real one needs a wooden base with grain, a steel strip, a brass terminal and a ruler with proper tick marks. Writing a shared helper for each of these (a wood texture, a steel material, a ruler, a glass tube, a spring) made later labs cheap to build and made all of them look like they belong to one lab bench. I now check every new scene against a photo of the real apparatus before I call it done.

Lesson 2: the physics must be right before the picture is pretty

Students will compare the readings with their textbook. Each lab has a small engine behind it: an Ohm's law lab gives real resistance values with a tolerance, a double-slit lab computes fringe width as lambda D / d, a Carnot engine lab follows the p-V cycle. I wrote unit tests for the formulas, so a visual change cannot quietly change a result.

Lesson 3: do not ship paid lab code to every visitor

The lab modules are not part of the website bundle. The API sends a lab's code only after it has checked the signed-in student's plan, so the free labs stay free and the paid ones are not sitting in a public JavaScript file.

Lesson 4: phones decide your scene budget

Most students open the labs on a phone. That meant fewer lights, small textures, and no heavy post-processing. I also made the loading screen identical for every lab, so waiting feels the same whether it is physics or chemistry.

Lesson 5: make the content findable

A great lab nobody can find is useless. Each lab has its own page with a title, a description, JSON-LD (LearningResource, FAQPage, BreadcrumbList), a sitemap entry, and chapter guides such as ray optics and current electricity that explain the formulas and link to the labs. I also publish an llms.txt so AI assistants can describe the site correctly.

What is next

More chemistry benches, a better teacher view, and more Hindi explanations. If you work on educational simulations I would like to hear how you handle realism versus performance. You can look at the labs here: practisci.in/labs.