How Does Calculus Integrate with Physics Concepts like Electric Fields and Flux?

[Josh930]

Homework Statement

Integration
I am soooo lost. I don't even know if this is the right forum... But where is the bridge between Calculus and Physics? I can Integrate equations, but when it comes to physics, i for one, don't know when to integrate; two, i don't see how you find the constants to remove from the integral; and three, Even given the integral formula for an equation, i still don't know what I am doing. ?? finding the electric field of an object?? i thought Electric field was (1/4pi*epsilon naught)(Q/r^2). So how do i find the E field for different shapes?

another example...

ex. Va-Vb=SE.dl

S-integral
E-Electric Field
dl-small segments of length

I don't know how to use the equation;

Or, electric flux,

Flux=SE . dA

What am i not understanding. Please help

Homework Equations

Flux=SE . dA

Va-Vb=SE.dl

The Attempt at a Solution

 

[tiny-tim]

Welcome to PF!

Hi Josh930! Welcome to PF!

(have an integral: ∫ and a pi: and an epsilon: ε and try using the X² and X₂ tags just above the Reply box )

… when it comes to physics, i for one, don't know when to integrate

Some physical quantities are A times B, or (vector) A dot B or A cross B …

for example, work done = force times distance …

if A and B are constant, then you just multiply, but if one or both is varying, then you have to integrate, eg: ∫A dB or ∫A.dB

i don't see how you find the constants to remove from the integral

Do you mean the constant of integration? You choose it to fit the initial (or boundary) condition: eg, you might choose potential energy to be zero at infinite distance.

finding the electric field of an object?? i thought Electric field was (1/4pi*epsilon naught)(Q/r^2). So how do i find the E field for different shapes?

another example...

ex. Va-Vb=SE.dl

S-integral
E-Electric Field
dl-small segments of length

I don't know how to use the equation;

Or, electric flux,

Flux=SE . dA

To find E at position x for different shapes, basically you integrate ∫∫∫ Q(r - x)d³r/4πε₀(r - x)³

V_a - V_b = ∫E.dl is the work-energy theorem: the LHS is the increase in PE, and the RHS is the work done … if E varies, then you need to integrate.

And yes, electric flux = ∫ E.dA … what is worrying you about that?