Question

A ring placed along y^2 + z^2 = 4, x = 0 carries a uniform charge of 5 uC/m. Find D at P(3,0,0).

Answer 1

What is "D"?

Answer 2

Electric Flux Density. Also using Coulombs laws. \[D = \epsilon _{0}*E\]

Answer 3

Electric flux as in
\[\huge \phi_e={q\over \epsilon_0}\]?

Answer 4

We haven't used an equation with that notation. does q represent electric field intensity? If so then yes i do believe we are talking about the same thing. \[D/\epsilon _{0} = E <-- electricfield intenisity \]

Answer 5

we have electric flux as \[\Psi = Q _{enc} (Charge enclosed) = \int\limits_{v}^{} \rho _{v} dv\] for example but D means electric flux density

Answer 6

ok your notation is ENTIRELY different from my own textbook so I'm somewhat lost

Answer 7

In my textbook coulomb's law is
\[\huge F = k_e{q_1 q_2 \over r^2}\]q represents charge

Answer 8

Alright we're still on the same page there!

Answer 9

so long as \[k _{e} = \frac{ 1 }{ 4piepsilon _{0} }\]

then I have electricfield intensity = \[\lim_{Q \rightarrow 0} \frac{ F }{ Q } \]

Such that:\[E =\frac{ Q (r - r')}{ 4\pi \epsilon _{0} |r-r'|^3}\]

Answer 10

okay and electric flux:
\[\huge \phi_e = \int \vec E \dot{d\vec A}\]

Answer 11

not quite... I have it as:

Answer 12

\[\Psi = \int\limits_{}^{} D ds = Q _{enc}\]

Answer 13

so you're looking for the electric field?
If my E is your D then E represents the electric field

Answer 14

Hmmm funny that they wouldn't use E but would use D to represent electric field. But okay. I am looking for electric flux density which is found by

Electric flux Density = \[\int\limits_{region}^{} \frac{ p_{r}dr }{ 4piR^2}\]

Answer 15

opps also multiply by the unit vector \[a _{r}\]

Answer 16

just a minor favor, since the text is so small, can you enlarge it using
\huge
?

Answer 17

ok have you learned Gauss's law?

Answer 18

okay i will in the future, we have but for this question it technically shouldn't be used yet. Gauss' law is the next section of questions that follows this one in the book titled 'gauss law and applications'.

Answer 19

take a look at the picture i attached. I feel like that is essentially what you're dealing with. But with all of the differences between our texts, it's hard to make sure we're on the same page so I just uploaded that from my text

Answer 20

hopefully it's what you need and makes SOME sense

Answer 21

got the answer! It took a couple more steps but you got 90% of the way there. Thanks!

Answer 22

sorry I couldn't help 100%
:)

Answer 23

You got me most of the way, more than i could have done!