Question

An infinite, nonconducting sheet has a surface charge density σ = +6.80 pC/m2. (a) How much work is done by the electric field due to the sheet if a particle of charge \(q_0 = 6.41 \times 10^{-19}\) C is moved from the sheet to a point P at distance d = 2.18 cm from the sheet? (b) If the electric potential V is defined to be zero on the sheet, what is V at P?

Answer 1

so first you have to use Gauss' law to get \(\vec E\), have you got that yet?

Answer 2

Sorry, I made some errors in the LaTeX above. If it is too confusing, you may need to refresh your page...

And no,

It is \(\large{\Phi=\int \vec E \cdot d\vec A}\)

Correct?

Answer 3

correct... technically a double surface integral, but I know you are overlooking those mathematical details

Answer 4

have you ever found the E-field due to an infinite plane of charge?
any idea what gaussian surface to choose?

Answer 5

For the class that I am in, Calculus III is technically not a prerequisite. I am in it now, but we have not gotten to the multiple integration or differentiation yet.

And, I don't believe that I have done that before. But if I had to guess, a cube?

Answer 6

a cube would work, actually :)
usually they call it a "pillbox" (which is kinda like a cylinder) but it actually works fine with any shape that has vertical sides

Answer 7

What is this sorcery that you speak of? Pillbox? I must learn it's ways...

Answer 8

ROLF @austinL

Answer 9

oops, spelled that wrong

Answer 10

hopefully it's clear that the E-field will be perpendicular to the plane, by symmetry.
the "pillbox" looks like this:

Answer 11

mind you that A need *not* be a circle; it can be any funky shape. YOu will see that A actually drops from the equation, so long as \(\vec A\) is either parallel or antiparallel with \(\vec E\)

Answer 12

Because it will on be both sides of the equation when you apply Gauss's Law

Answer 13

Okay, I am following along me thinks...

Answer 14

it will make more sense when you actually do gauss law here
and use the form \(\large\frac q{\epsilon_0}=...\)

Answer 15

Keep in mind that, based on @TuringTest 's drawing, you have to double the E-field since the gaussian surface is both above and below the plane of charge

Answer 16

I'm sorry, my brain is experiencing the mid-afternoon fuzzies... I am afraid I am having some difficulty grasping this.

Answer 17

\[\huge \phi_e = 2EA = {\sigma A \over \epsilon_0 }\]
\[\huge \phi_e\] is the electric flux and the charge enclosed is the area A times the surface charge density sigma

Answer 18

But... the area is on an infinite sheet...

Answer 19

but we only need consider the area that has *enclosed* charge in it

Answer 20

what is the charge on this part of the sheet? (no, I don't want a numerical answer, in symbols please)