Is an electric field always changing?

[jaredvert]

With respect to 1/r^2 since the force decreases ? What does a uniform electric field mean then?

 

[Simon Bridge]

A uniform electric field means that the force does not depend on position.
The inverse-square law is only for a spherically symmetric charge distribution - you can arrange the charges into other shapes.

 

[jaredvert]

A uniform electric field means that the force does not depend on position.
The inverse-square law is only for a spherically symmetric charge distribution - you can arrange the charges into other shapes.

So you mean point particles or spheres right? So a parallel plate wouldn't be 1/r^2 dependence right?

 

[Simon Bridge]

That is correct - at long ranges quite a few things look like spheres.

A long line of charges has a field that falls off as 1/r and a flat sheet has a constant field that switches sign at the sheet position. 2 parallel plates, opposite charge, have a uniform field between the plates but zero outside... and so on.

Fields can get much more complicated.
The field due to two or more charged spheres does different things in different directions.
... and there are more shapes than those to choose from.

If you are doing an electrostatics course you will get to calculate some of the fields for simpler distributions of charge.

 

[jaredvert]

That is correct - at long ranges quite a few things look like spheres.

A long line of charges has a field that falls off as 1/r and a flat sheet has a constant field that switches sign at the sheet position. 2 parallel plates, opposite charge, have a uniform field between the plates but zero outside... and so on.

Fields can get much more complicated.
The field due to two or more charged spheres does different things in different directions.
... and there are more shapes than those to choose from.

If you are doing an electrostatics course you will get to calculate some of the fields for simpler distributions of charge.

Yeah I'm in high school physics and now that I think about it we derived the equations for these and i do remember two parallel plates not depending on r at all. That makes perfect sense! Thank you. I was also hoping. You could elucidate the topic of voltage as well? In my book it says a conducting sphere has voltage equal to 4pi epsilon times 1/radius of sphere relative to infinity. Well I get the infinity part but I'm wondering why the generic "r" is equal to the radius of the sphere? I mean wouldn't it have to be infinity minus r (which is just infinity and therefore makes the voltage 0?). Basically if you take it relative to infinity and you know of no other charges then how does it have voltage? Thanks for the help

 

[Simon Bridge]

For a conducting sphere radius R and charge Q:

- when $r\greq R$ the voltage is $$V=\frac{1}{4\pi\epsilon_0 r}$$

- when $r<R$ the voltage is $$V=0##

That help?

The voltage is the potential energy per unit charge.

The potential energy stored in an object is the amount of work you have to do to assemble the object from a condition where none of the bits exert any forces on each other.

So you can imagine that all the little charges that make up the sphere - the only place they can have zero force on each other is for them all to be an infinite distance apart. Moving the same charges closer together than that requires work to overcome their mutual repulsion. That is how we know it has a voltage.

 

[jaredvert]

Yeah dude thanks for the help broski !

 

[Simon Bridge]

No worries.
Enjoy.