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Let's take into consideration a system of two charges, with magnitudes $q_1$ and $q_2$.Let's assume that the charge with magnitude $q_1$ is placed at the point $A$ and also assume that a point $B$ is at the distance of $r_1$ from $A$ and another point $C$ is at a distance of $r_2$ from $A$. Points $A$, $B$ and $C$ are collinear. The diagram below will make the above statements clearer.

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The charge $q_2$ will further be moved to the point $B$ from $C$ The thing that I was mainly wondering about was why we don't take a calculus-based approach when evaluating something like the work done when we push/pull an object of a certain mass, with a certain force, over a certain distance.
That's when I thought that the force between $q_1$ and $q_2$ is changing with the change in position of $q_2$, unlike the situation described above, where the force applied by us is constant.
[Note : I do know that the force applied by us while pushing or pulling something may not be constant but I say this just because that's the situation in most of the examples that I'm familiar with]

Here's what the derivation for work done when $q_2$ moves from $C$ to $B$ is given in my textbook :
Let's assume that the charge $q_2$ suffers a tiny displacement $dr$, so the work done will be $F.dr = \dfrac{q_1q_2}{4\pi\varepsilon_0r_2^{\text{ }2}}dr$.
Hence, the total work done will be the sum of all these little work(s) done as $q_2$ moves from $C$ to $B$ which can be written as $$W = \int_{r_2}^{r_1}\dfrac{q_1q_2}{4\pi\varepsilon_0r^2}dr$$ The above stated expression is further simplified as $\dfrac{q_1q_2}{4\pi\varepsilon_0}\Bigg ( \dfrac{1}{r_1}-\dfrac{1}{r_2} \Bigg )$.

Is the idea that we approach this problem using calculus because of the continuous change in force as $q_2$ changes its position correct?
Pardon me if this is something obvious because I'm an absolute beginner to calculus.

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    $\begingroup$ If I may ask, how have you concluded that the work done due to pushing/pulling is not calculated by calculus? $\endgroup$ Jun 15, 2020 at 7:49
  • $\begingroup$ I'm in $10^{th}$ grade and mechanics was a major part of $9^{th}$ grade Physics and all of the examples there included the application of a constant force, so I concluded that. I do know that its not necessary that a constant force is applied while pulling or pushing something but the only reason I stated that was because that was the case with most of the situations I was familiar with. Also, this is the first time I saw the use of calculus in evaluating the work done. $\endgroup$ Jun 15, 2020 at 7:58
  • $\begingroup$ In that case, I suggest that you look up a High School textbook(Year 11/Year 12) and in that you may find what you are looking for. In general, if there is a constant force acting on the object, the work done on it may indeed be calculated by calculus, specifically - by integration. If you're an Indian High School Student, some of the text books that I have seen have very good examples to explain this concept. $\endgroup$ Jun 15, 2020 at 8:02
  • $\begingroup$ Thanks! I would really appreciate it if you could give me an example of a situation like that so I know what I'm looking for. Thanks, again :) $\endgroup$ Jun 15, 2020 at 8:04
  • $\begingroup$ Are you looking for a physical example or a mathematical example will do? $\endgroup$ Jun 15, 2020 at 8:05

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Well, this is an extremely simple example to elucidate what I had already said in the comments.

First off, Work is defined as: $$ W = \vec F. \vec d$$ where, $\vec F$ is the force applied and $\vec d$ is the displacement. We have taken the dot product of $\vec F$ & $\vec d$ because Work is a scalar quantity. It can be simplified as $$ W = Fd\cos{\theta}$$ where, $\theta$ is the angle between $\vec F$ and $\vec d$.

A simple yet intuitive example is when a rain drop falls down to the earth. You may be aware that a raindrop falls down because of the downward gravitational force applied by the earth.

So, if a raindrop fall down from a height $h$ and has a mass $m$, the work done by the earth on the raindrop will simply be: $$ W = \vec F. \vec d$$

Here, $\vec F = mg$, where $g$ is the acceleration due to gravity. Also, if you notice, the displacement (i.e. $h$) and the force acting on the rain drop (i.e. $mg$) are acting along the same direction (i.e. downwards). So, the angle between them will be $0$, and $cos\ 0 = 1$. So, the work done reduces to

$$W = (mg).(h).cos\ 0$$ i.e. $$W = mgh$$

Hope that you understood this example!

Note: This requires a primitive understanding of vectors and and their products

Edit: Calculus based example:

Suppose there is a woman pushing a block. She applies a constant force of $\vec F = 50\ N$. Now, the block moves along a smooth, horizontal surface. Let's set a reference axis now. Let the horizontal surface be the x - axis. Now, let's say the block moves a distance of $2\ m$ from the time of application of the constant force.

If we consider the origin to be at the point where the block initially is, the work done would simple be written as: $$\int dW\ = \int_{x=0}^{x=2} F. dx.cos\ 0$$ where, $dx$ is an infinitesimal displacement

Again, the angle between the force and the displacement is zero and $F$ is constant, so they can be taken out of the integration. Also, the integration of $dx$ is simply $x$.

So it simplifies as: $$W\ = 20\ [2-0]$$ Hence, $W=\ 40\ J$

Hope that this served the purpose!

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  • $\begingroup$ Thanks! But I don't see calculus being used in this example. Wasn't that supposed to be included in the example that you provide? $\endgroup$ Jun 15, 2020 at 8:34
  • $\begingroup$ Oops! Sorry! I'll add that too! Sorry! $\endgroup$ Jun 15, 2020 at 8:35
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    $\begingroup$ Look, calculus is the most fundamental way to approach a problem like this. All other methods are not really 'methods' in the sense, all of them come from calculus! In that case, the other methods are just the final step simplifications that are arrived at by calculus. Even in your post, the 'alternative' method you have described is the final step of the calculus-based derivation. So, for simple shortcuts like those, just remember the formula, $W=\vec F.\vec d$. $\endgroup$ Jun 15, 2020 at 9:08
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    $\begingroup$ No problem! Glad to help! It's really nice to see someone at a young age really demonstrating that they really want to learn something from this site and not just use it for homework help!! $\endgroup$ Jun 15, 2020 at 9:16
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    $\begingroup$ Thank You for the compliment :) $\endgroup$ Jun 15, 2020 at 9:18

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