Derivatives in physics

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  • #1
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Can anyone explain what actually use of derivatives in physics.It's totally beyond my understanding.I was doing gauss law and i came across this derivative doubt.In the video at time 8:13 to 8:33


what he means by saying if area is small electric field should be approximately constant?is he talking about uniform electric field?Or he means all such dA would have same value and direction of electric field?
what he actually meant by constant electric field -
(1)same electric field for all dA OR
(2)uniform electric field?
which one ( 1)or (2)?
 
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  • #2
DrClaude
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This is not a derivative. He is considering an infinitely small area dA. To this infinitely small area corresponds an infinitely small flux dΦ. To get the total flux Φ, you need to sum the small fluxes, but since the fluxes are infinitely small, the sum becomes an integral.

I think the explanation given in the video is pretty good for this level. Can you point out more specifically what you don't understand?
 
  • #3
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Twice, he said "the electric field is approximately constant over it", it being the small element of area dA. Even if the electric field is varying on the large scale, over a small enough area, it doesn't vary much. That's all he was trying to say.

Chet
 
  • #4
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Even if the electric field is varying on the large scale, over a small enough area, it doesn't vary much. That's all he was trying to say.
how electric field vary?what is variation in electric field?does variation in electric field mean electric field is not uniform ?
 
  • #5
jtbell
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how electric field vary?

It depends on the situation. For example, for a point charge, the magnitude (strength) of the field varies with distance r from the charge: $$E = \frac{q}{4 \pi \epsilon_0 r^2}$$ (Coulomb's Law). The direction of the field varies with location because it's always directly away from a positive charge, or directly towards a negative charge.
 
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  • #6
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It depends on the situation. For example, for a point charge, the magnitude (strength) of the field varies with distance r from the charge: $$E = \frac{q}{4 \pi \epsilon_0 r^2}$$ (Coulomb's Law). The direction of the field varies with location because it's always directly away from a positive charge, or directly towards a negative charge.
So how does infinitely small area REDUCES ELECTRIC FIELD VARIATION?
 
  • #7
jtbell
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To make an analogy, the earth is a sphere, so the direction of "up" (away from the center of the earth) varies from one location to another. It's much different in London than in Singapore. :D But if you stay inside London, "up" is very nearly the same everywhere.

Likewise, the strength of the earth's gravitational field varies as 1/r2 where r is the distance from the center of the earth. Nevertheless, in typical laboratory situations near the earth's surface we usually assume it's uniform with height (9.81 m/s2). The difference between floor and ceiling is far too small to bother with.
 
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  • #8
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To make an analogy, the earth is a sphere, so the direction of "up" (away from the center of the earth) varies from one location to another. It's much different in London than in Singapore. :D But if you stay inside London, "up" is very nearly the same everywhere.

Likewise, the strength of the earth's gravitational field varies as 1/r2 where r is the distance from the center of the earth. Nevertheless, in typical laboratory situations near the earth's surface we usually assume it's uniform with height (9.81 m/s2). The difference between floor and ceiling is far too small to bother with.
hmmm.........great analogy .So here earth is total area say A.direction of up is electric field.London ,singapore and several other countries would be different dA .
 
  • #9
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To make an analogy, the earth is a sphere, so the direction of "up" (away from the center of the earth) varies from one location to another. It's much different in London than in Singapore. :D But if you stay inside London, "up" is very nearly the same everywhere.

Likewise, the strength of the earth's gravitational field varies as 1/r2 where r is the distance from the center of the earth. Nevertheless, in typical laboratory situations near the earth's surface we usually assume it's uniform with height (9.81 m/s2). The difference between floor and ceiling is far too small to bother with.
But what about integrating all those dA?won't it become A again so again variable electric field at different points?
 
  • #10
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To make an analogy, the earth is a sphere, so the direction of "up" (away from the center of the earth) varies from one location to another. It's much different in London than in Singapore. :D But if you stay inside London, "up" is very nearly the same everywhere.

Likewise, the strength of the earth's gravitational field varies as 1/r2 where r is the distance from the center of the earth. Nevertheless, in typical laboratory situations near the earth's surface we usually assume it's uniform with height (9.81 m/s2). The difference between floor and ceiling is far too small to bother with.
Please jtbell reply.I am waiting for your magical analogy and explanation.
 
  • #11
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But what about integrating all those dA?won't it become A again so again variable electric field at different points?
No. You are integrating over an area, but the integrand of dA is not 1. In the case being considered in the video, the integrand is the component of the field strength normal to dA. So, of course, you don't get A when you integrate, since the component of the field strength normal to dA varies over the surface.

Chet
 
  • #12
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Till dA i understood but why integration is done ?
 
  • #13
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To make an analogy, the earth is a sphere, so the direction of "up" (away from the center of the earth) varies from one location to another. It's much different in London than in Singapore. :D But if you stay inside London, "up" is very nearly the same everywhere.

Likewise, the strength of the earth's gravitational field varies as 1/r2 where r is the distance from the center of the earth. Nevertheless, in typical laboratory situations near the earth's surface we usually assume it's uniform with height (9.81 m/s2). The difference between floor and ceiling is far too small to bother with.
Till dA i understood but why integration is done ?
 
  • #14
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Till dA i understood but why integration is done ?
In the example on the video, the guy is trying to determine the "electric flux" through a surface. This requires you to do integration over the surface if the normal component of the field strength is varying with position at the surface.

Chet
 
  • #15
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Here is my understanding.We want flux i.e lines of electric field E linked with an area A.So flux is proportional to area as well as electric field.So we can multiply electric field in unit area ( if it is same throughout the area)) and multiply it by area A.But what if electric field is not uniform?So for this we take extremely small area dA and being so small electric field here is constant (does not vary) and we will take what is electric field in this dA .Then we multiply this electric field( electric field in dA)with integral (i.e adding up all such dA)to get total flux.
Where i am wrong?
here i have assumed electric field lines to be straight not at any angle.
 
  • #16
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Here is my understanding.We want flux i.e lines of electric field E linked with an area A.So flux is proportional to area as well as electric field.So we can multiply electric field in unit area ( if it is same throughout the area)) and multiply it by area A.But what if electric field is not uniform?So for this we take extremely small area dA and being so small electric field here is constant (does not vary) and we will take what is electric field in this dA .Then we multiply this electric field( electric field in dA)with integral (i.e adding up all such dA)to get total flux.
Where i am wrong?
here i have assumed electric field lines to be straight not at any angle.
Yes, if you mean by the last sentence that you are describing the special case in which the field lines are perpendicular to the surface at each location on the surface.
 
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  • #17
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So we can multiply electric field in unit area ( if it is same throughout the area)) and multiply it by area A to get the flux
Is it correct?
 
  • #19
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( electric field in dA)with integral (i.e adding up all such dA
What integral of dA would be in this case?Total area A ,right?
 
  • #21
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O
Sure.

Chet
OK.One more question.We should multiply electric field in unit area with total area to get flux but here for a reason that electric field was not uniform we took electirc field in dA Area( which is surely smaller than unit area )and multiplied it with total area A .so how can this give correct result?
 
  • #22
Doc Al
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What integral of dA would be in this case?Total area A ,right?
To find the total flux through an area, you want the integral of E dA, not just dA. That makes a difference when E varies over the surface.

Of course, if E is constant over the surface, you can pull it out of the integral and just integrate dA, which is the total area.
 
  • #23
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To find the total flux through an area, you want the integral of E dA, not just dA. That makes a difference when E varies over the surface.
But that's why we took dA instead of A so that we get constant electric field so every time we take dA we can always put E out of integral?
 
  • #24
Doc Al
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But that's why we took dA instead of A so that we get constant electric field so every time we take dA we can always put E out of integral?
##\int E dA = E \int dA = EA## when E is constant (and normal to the surface).
 
  • #25
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so every time we take dA we can always put E out of integral?
according to 7th post of jetbell.please answer.
 
  • #26
DrClaude
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But that's why we took dA instead of A so that we get constant electric field so every time we take dA we can always put E out of integral?
No, you take dA such that E is constant over dA, but E is not constant over all A.
 
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  • #27
jtbell
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Think of the integral as being a sum of a lot of pieces, one for each small part of the surface: $$\int {\vec E \cdot d \vec A} = \vec E_1 \cdot d \vec A_1 + \vec E_2 \cdot d \vec A_2 + \cdots$$ Suppose you have a point charge q and make the surface a sphere, with the charge at its center. Then, first, ##\vec E## is perpendicular to the surface at all points on the sphere, so the vector "dot product" becomes an ordinary product of the magnitudes: $$\int {\vec E \cdot d \vec A} = E_1 dA_1 + E_2 dA_2 + \cdots$$ Second, ##\vec E## has the same magnitude at all points on the sphere, so ##E_1 = E_2 = \cdots##, let's call it just E: $$\int {\vec E \cdot d \vec A} = E (dA_1 + dA_2 + \cdots)$$ Now, the sum of all the dA's is just the total surface area of the sphere, which is... (your turn)

In this case, the integration is easy because it's only a "conceptual" integration, so you don't need the complicated methods that you learn in a calculus course.

Now, suppose we put the charge at the center of a cube instead of a sphere. We can still calculate ##\int {\vec E \cdot d \vec A}## directly, but it's much more complicated because the magnitude of ##\vec E## is different at different points on the cube surface, and ##\vec E## has different angles to the surface at different points on the cube. I think the last time I did it, it took me about two pages of calculus and algebra. But, in the end we get the same result as with a sphere! In fact, no matter what shape the surface is, so long as it is a closed surface with the charge inside it, we get the same result! The most general statement of this is Gauss's Law: $$\int {\vec E \cdot d \vec A} = \frac{q}{\epsilon_0}$$
 
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  • #28
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Think of the integral as being a sum of a lot of pieces, one for each small part of the surface: $$\int {\vec E \cdot d \vec A} = \vec E_1 \cdot d \vec A_1 + \vec E_2 \cdot d \vec A_2 + \cdots$$ Suppose you have a point charge q and make the surface a sphere, with the charge at its center. Then, first, ##\vec E## is perpendicular to the surface at all points on the sphere, so the vector "dot product" becomes an ordinary product of the magnitudes: $$\int {\vec E \cdot d \vec A} = E_1 dA_1 + E_2 dA_2 + \cdots$$ Second, ##\vec E## has the same magnitude at all points on the sphere, so ##E_1 = E_2 = \cdots##, let's call it just E: $$\int {\vec E \cdot d \vec A} = E (dA_1 + dA_2 + \cdots)$$ Now, the sum of all the dA's is just the total surface area of the sphere, which is... (your turn)

In this case, the integration is easy because it's only a "conceptual" integration, so you don't need the complicated methods that you learn in a calculus course.

Now, suppose we put the charge at the center of a cube instead of a sphere. We can still calculate ##\int {\vec E \cdot d \vec A}## directly, but it's much more complicated because the magnitude of ##\vec E## is different at different points on the cube surface, and ##\vec E## has different angles to the surface at different points on the cube. I think the last time I did it, it took me about two pages of calculus and algebra. But, in the end we get the same result as with a sphere! In fact, no matter what shape the surface is, so long as it is a closed surface with the charge inside it, we get the same result! The most general statement of this is Gauss's Law: $$\int {\vec E \cdot d \vec A} = \frac{q}{\epsilon_0}$$
it means in case of sphere,electric field is same for all points on the surface of sphere ,so all dA,unit area and the complete area A all have same electric field.
 
  • #29
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Now, the sum of all the dA's is just the total surface area of the sphere, which is... (your turn)
4 pi r^2 where r=radius of sphere
 
  • #30
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No, you take dA such that E is constant over dA, but E is not constant over all A.
But in case of sphere E is constant over dA, as well as total area A,right?
 

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