Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Possible Duplicate:
Why are continuum fluid mechanics accurate when constituents are discrete objects of finite size?

When we apply differentiation on charge being conducted with respect to time,i.e dq/dt, we consider the charge flown to be infinitely small, but q cannot be less than 1.602*10^-19. So how can we assume this to be infinitely small?

share|improve this question

marked as duplicate by Qmechanic, Manishearth Dec 10 '12 at 9:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Possible duplicate: physics.stackexchange.com/q/21051/2451 –  Qmechanic Aug 14 '12 at 16:29

2 Answers 2

up vote 2 down vote accepted

Even a physical quantity which changes by discrete amounts can often be well approximated by a continuous function of time.

The derivative is a property of a mathematical function. Any differentiable function must necessarily be continuous, and a continuous function will change by arbitrarily small values for an arbitrarily small change in inputs.

The fact that one can calculate the derivative of a function does not imply that the physical quantity that is approximated by that function can also be changed by arbitrarily small amounts.

share|improve this answer

Physics is all about making the right approximations, in the hope that we can gain some actual physical insight into our problem and make verifiable predictions.

For example, say you wanted to calculate the trajectory of a cannonball that has been fired from a cannon. It would be a Sisyphean task to account for all the possible variables that could affect the motion of the cannonball: air resistance, wind, recoil of the wheels of the cannon, rotation of the ball, blemishes on the ball's surface, lift, curvature of the Earth... It's pretty much impossible to come up with a full list of things that could affect it, let alone factor those into your calculation!

But fortunately, these effects are far too small to have a real effect on whether the ball hits the target or not. So you're safe make approximations - assume the Earth is locally flat, assume you're in a vacuum, assume the ball is a sphere, and so on, in the hope of making your problem tractable. We all agree that the cannonball doesn't travel in a perfect parabola, as your sums predict, but it's pretty damn close, and we've got a grip on the interesting parts of the problem.

Aside: How do you decide what a small enough effect to ignore is? Well, we've said nothing of the fact that your instruments are unreliable too: perhaps you hit the stopwatch a little early or late, or your ruler changed size a little in the heat of the Sun, or you couldn't quite locate the centre of the hole the cannonball made in the ground. You admit that your measurement is inherently inaccurate, as all measurements are, and make an estimate of the uncertainty in your answer. You're damn well never going to see the effect of air resistance on the cannonball if it's smaller than your uncertainty. This is why scientists put so much effort into devising ever more precise instruments - so we can measure physics on smaller and smaller scales.

So you see, the assumption in classical electrodynamics that charge is a continuous variable is just an approximation, just like assuming the cannonball is a sphere. We all know that charges come in quanta of $e$, but if we're measuring a charge of 10 Coulombs - well, who cares about the electrons?

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.