What Is the Interpretation of an Integral Equation with a Proportional Variable?

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SUMMARY

The discussion centers on the interpretation of the integral equation $$\Delta V = - \int_a^b \vec E \cdot d \vec s$$, which represents the change in electric potential between two points in an electric field. The equation is derived from the concept of summing small changes in potential, represented as $$\delta V = -\vec{E} \cdot \delta \vec{r}$$, and transitioning from discrete sums to continuous integrals. Participants emphasized that this integral captures the cumulative effect of the electric field along a path, providing a clear interpretation of potential difference.

PREREQUISITES
  • Understanding of integral calculus, particularly line integrals.
  • Familiarity with vector fields and electric fields.
  • Knowledge of electric potential and its relationship to work done by electric forces.
  • Basic concepts of limits and summation processes in calculus.
NEXT STEPS
  • Study the concept of line integrals in vector calculus.
  • Explore the relationship between electric fields and work done, specifically $$W = \int_a^b \vec F \cdot d \vec r$$.
  • Investigate the mathematical foundations of potential energy in electric fields.
  • Learn about the applications of integral equations in physics, particularly in electromagnetism.
USEFUL FOR

Students of physics, electrical engineers, and anyone interested in understanding the mathematical foundations of electric potential and field interactions.

Anti Hydrogen
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Homework Statement
Hi, I'm trying to understand the case when a variable is proportional to that which is under a integral sign, for example
$$\Delta V = - \int_a^b \vec E \cdot d \vec s$$

What could be a interpretation of this equation?
Relevant Equations
Thanks
Thanks
 
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Anti Hydrogen said:
Homework Statement:: Hi, I'm trying to understand the case when a variable is proportional to that which is under a integral sign, for example
$$\Delta V = - \int_a^b \vec E \cdot d \vec s$$

What could be a interpretation of this equation?
Relevant Equations:: Thanks

Thanks
An integral is the limit of a summation process. Here we have ##\Delta V## as (the sum of) the product of the electric field multiplied by a length. You could compare this with:
$$W = \int_a^b \vec F \cdot d \vec r$$
 
Essentially this is a restatement what @PeroK was saying: consider the change in the potential ##\delta V## along a small displacement vector ##\delta \vec{r}## in a region where the electric field is ##\vec{E}##, $$\delta V = -\vec{E} \cdot \delta \vec{r}$$Now if we sum up all of these small changes along some path $$\sum \delta V = -\sum \vec{E} \cdot \delta \vec{r}$$If you now take the limit as the length of the displacement vector approaches zero, that is ##\delta \vec{r} \rightarrow d\vec{r}## and consequently also ##\delta V \rightarrow dV##, the sums turn into integrals $$\int_{V_{1}}^{V_{2}} dV = -\int_{a}^{b} \vec{E} \cdot d\vec{r}$$Apologies if I abused any notation! For this specific integral, the interpretation is the change in potential going from ##a## to ##b##.
 
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Anti Hydrogen said:
Homework Statement:: Hi, I'm trying to understand the case when a variable is proportional to that which is under a integral sign, for example
$$\Delta V = - \int_a^b \vec E \cdot d \vec s$$

What could be a interpretation of this equation?
Relevant Equations:: Thanks

Thanks
What variable is proportional to what under the integral sign? Are you saying that ##\Delta V## is proportional to ##\vec E \cdot d \vec s?## @PeroK said it right, look at the line integral as a summation.
 

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