Electric Field

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bananaburrito
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Why is electric field strength equal to the negative potential gradient?
I've read through many websites but none of the explanations helped me to really understand this concept.
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MouldyVinegar
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The area under a force - distance graph is the change in potential energy, or in other words the integral or antiderivative of force with respect to distance is the total change in potential energy.

Therefore, the derivate of the potential energy undoes the integral or antiderivative, and thus returns the force.

 \int F(x) dx = P(x)
Where F(x) and P(x) is the force and potential at a distance x respectively
 \frac{d}{dx} P(x) = \frac{d}{dx} \int F(x) dx = F(x)

The reason behind the negative sign is due to how the derivate is computed. What is the derivative of  1/x  with respect to x?

And thus what is the derivative of potential energy with respect to distance? (Remember that often r is used to show distance)

Unfortunately, the explanation is quite mathsy, and if you don't do maths at A-Level I don't think it is possible to explain it, (although I'm not the greatest at physics, so who knows? Maybe there is a much more intuitive way to explain)
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Eimmanuel
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(Original post by bananaburrito)
Why is electric field strength equal to the negative potential gradient?
I've read through many websites but none of the explanations helped me to really understand this concept.

Do you mind quoting the websites that you have come across so that we can know what are the explanations that do not help you instead of just saying that you have read through many websites?

The following explains the “why” in an indirect way.
https://openstax.org/books/universit...from-potential
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Eimmanuel
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(Original post by bananaburrito)
Why is electric field strength equal to the negative potential gradient?
I've read through many websites but none of the explanations helped me to really understand this concept.
I disagree with what MouldyVingear had explained.
“The reason behind the negative sign is due to how the derivative is computed.”
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Eimmanuel
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(Original post by MouldyVinegar)
The area under a force - distance graph is the change in potential energy, or in other words the integral or antiderivative of force with respect to distance is the total change in potential energy.

Therefore, the derivate of the potential energy undoes the integral or antiderivative, and thus returns the force.

 \int F(x) dx = P(x)
Where F(x) and P(x) is the force and potential at a distance x respectively
 \frac{d}{dx} P(x) = \frac{d}{dx} \int F(x) dx = F(x)

The reason behind the negative sign is due to how the derivate is computed. What is the derivative of  1/x  with respect to x?

And thus what is the derivative of potential energy with respect to distance? (Remember that often r is used to show distance)

Unfortunately, the explanation is quite mathsy, and if you don't do maths at A-Level I don't think it is possible to explain it, (although I'm not the greatest at physics, so who knows? Maybe there is a much more intuitive way to explain)

Note that OP is asking “Why is electric field strength equal to the negative potential gradient?”, the relationship between electric field strength and potential gradient NOT electric force and the gradient of potential energy.

Although the relationship between electric force and the gradient of potential energy is similar to the relationship between electric field strength and potential gradient, you still need to explain how can one arrive at the relationship between electric field strength and potential gradient from the relationship between electric force and the gradient of potential energy.


(Original post by MouldyVinegar)
...The reason behind the negative sign is due to how the derivate is computed. What is the derivative of  1/x  with respect to x? ...
Do you have a reference for this or this is how you have understood the minus sign?
If this is coming from your teacher, I suggest you request the source.
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MouldyVinegar
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(Original post by Eimmanuel)
Note that OP is asking “Why is electric field strength equal to the negative potential gradient?”, the relationship between electric field strength and potential gradient NOT electric force and the gradient of potential energy.

Although the relationship between electric force and the gradient of potential energy is similar to the relationship between electric field strength and potential gradient, you still need to explain how can one arrive at the relationship between electric field strength and potential gradient from the relationship between electric force and the gradient of potential energy.



Do you have a reference for this or this is how you have understood the minus sign?
If this is coming from your teacher, I suggest you request the source.
My understanding was that, for a point charge / mass / object,
 Potential = k/x  where k is some constant such as  \frac{Q}{4 \pi \epsilon _{0}} or  MG , and  x is the distance between the object and source of the force field.
 \frac{d}{dx} k/x = -k/x^{2} which is exactly of the form of the force acting on an object per unit of charge / mass / other unit. Furthermore, a minus sign is introduced here when the derivative is computed which also explains why it is equal to the negative of the gradient.

Is there any other mistakes / misconceptions in here?
Thanks for pointing out my mistakes though, the last thing I would want to do is mislead someone!
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Eimmanuel
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(Original post by MouldyVinegar)
My understanding was that, for a point charge / mass / object,
 Potential = k/x  where k is some constant such as  \frac{Q}{4 \pi \epsilon _{0}} or  MG , and  x is the distance between the object and source of the force field.
 \frac{d}{dx} k/x = -k/x^{2} which is exactly of the form of the force acting on an object per unit of charge / mass / other unit. Furthermore, a minus sign is introduced here when the derivative is computed which also explains why it is equal to the negative of the gradient.

Are there any other mistakes / misconceptions in here?
Thanks for pointing out my mistakes though, the last thing I would want to do is misled someone!
It seems that you are using a “circular” explanation for explaining the negative sign in the potential gradient.
Electric field strength equal to the negative potential gradient is applicable in all valid problems.

You are making use of a point charge system to “show” that electric field strength equal to the potential gradient and then say that you find the reason why there is a negative sign. This is not showing at all!

You are doing Example 7.17 in a backward manner.

I would illustrate what you are doing with an analogy that I can think of right now. Disclaimer: all analogies are wrong imo.

Say you want “prove” that the following differentiation “rule”:
 \dfrac{d}{dx} x^n = n x^{n-1}  for any integer of n.

Apply the rule to say  x^2 and you have
 \dfrac{d}{dx} x^2 = 2x

So the rule is found and is valid for any integer of n.
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