Question

In electric fields..why is \(d\vec E = \frac{k\text d q}{r^2} (\hat r)\)

Answer 1

you can often treat a charge distribution as a summation of infinitesimal point charges.

Answer 2

so the field is the summation of their individual fields, by superposition.

Answer 3

is that what you're asking?

Answer 4

no. i meant where did that formula come from?

Answer 5

Coulomb.

Answer 6

isn't coulumb's law \[| \vec F | = \frac {kqq'}{r^2} (\hat r)\]
how does that become dq?

Answer 7

qE =F or F/q =E , whichever you prefer: if there's no charge at some location in a field, there's still a field. as for the 'dq', I went over that above.

Answer 8

wait..where dod you get that?

Answer 9

E, in this definition, is something like "the field that causes a force, F, on a charge q', located a distance r from the charge generating the field, q. "

Answer 10

and q is the charge?

Answer 11

yep.

Answer 12

but that still doesn't explain why \[d\vec E = \frac {kdq}{r^2} (\hat r)\]

Answer 13

does it explain why E = kq/r^2 ?

Answer 14

because if it explains that, then it explains the differential form of that, since they're exactly the same but for scale.

Answer 15

if i remember right.. \[\huge \vec E = \frac{\left ( \frac{kqq_o}{r^2} \right)}{q_o} \implies \frac{kq}{r^2}\]
right?

Answer 16

yep.

Answer 17

now what?

Answer 18

now make the charge tiny.

Answer 19

the field is proportional to the charge.

Answer 20

So you must make its field scale correspondingly.

Answer 21

what do you mean make the charge tiny?

Answer 22

do you mean take its derivative?

Answer 23

x. dx. what's the difference?

Answer 24

dx is the derivative of x?

Answer 25

sort of. it's the infinitesimal of x. the differential form. if something is occurring on x, like say a Force, and everywhere on x that force is a bit different, then F(x)dx is the way you express it. F(x)x isn't going to usefully describe the situation.

Answer 26

hmm so why only take the d of q and not others?

Answer 27

k doesn't change.
r you have to write as either constant or as a function of something.
eg. for a line of charge along x, with a linear charge density of lambda, dq = lambda*dx
and hopefully you can also express the distance (and direction) from each of those little 'dq's' to the point where you want to know the value and direction of the field in terms of x also...

Answer 28

and q is just a variable?

Answer 29

that's why it's the only one that's put with a d?

Answer 30

it's a quantity.

Answer 31

here:

Answer 32

what do you mean by "a quantity"