Gauss' law for concentric circles

[gracy]

1/4πε0.(-2q )c^2

 

[ehild]

But in my graph #19 it has been given magnitude of 1/4πε0.-2q c^2

If something is written by you it need not be right.
Your formula is wrong. Use parentheses.
E is something at one side of the boundary and zero at the other side. You can calculate the limits of E at c at both sides, but they are not equal.

 

[gracy]

If something is written by you it need not be right.

It's in my book.Not by me.

 

[gracy]

I don't know how to right mathematical equations/formula here.Don't know about latex.

 

[ehild]

It can not be so in your book. You wrote $\frac{1}{4} π ε_0 (-2q) c^2$. Is the electric field proportional to the square of the radius?

 

[gracy]

$1/4πε0.(-2q)/c^2$

 

[ehild]

I don't know how to right mathematical equations/formula here.Don't know about latex.

Use parentheses.

 

[ehild]

$1/4πε0.(-2q)/c^2$

better. But πε0 is still at the wrong place. You should divide by it.

 

[gracy]

But πε0 is still in the numerator.

But I think (can see) it is denominator .

 

[ehild]

But I think (can see) it is denominator .

It is not. 1/ab means $\frac{1}{a}b$

 

[ehild]

It is not. 1/ab means $\frac{1}{a}b$

try to input to your calculator 10/2*5

 

[gracy]

$1πε0/4.(-2q)c^2$

 

[ehild]

$1πε0/4.(-2q)c^2$

I give up.

 

[gracy]

I give up.

Well,thanks for the answers you have given me so far.

 

[gracy]

@Doc Al have you also given up?Want to ask a question.

 

[gracy]

I now understand that electric field just inside the conductor is zero and just outside the conductor it is $σ/ε0$ so at at the r=c electric field has two values zero & $σ/ε0$ .As the graph in post 19 shows.
Right?

 

[BvU]

Ehild has been so patient with you. You really should repair that !

I notice that in the figure in post #19 (also a figure you reconstructed from what you have in the book ?) the same problem exists.

The expressions there should read, e.g. for the region b < r < c :$$E = {1\over 4\pi\epsilon_0}\, {-2q\ \over r^2} $$
Elizabeth really bent over backwards to make that clear to you.

In your post #136 you are either derailing badly or saying the right thing, depending on choices of $\sigma$ and the direction of $\vec E$. At the risk of triggering another 100+ posts:

If we take the r+ direction as positive, and proceed from 0 to b, the field is 0.

At b there is a jump from 0 to $\sigma\over \epsilon_0$. Here $\sigma ={ -2q\ \over 4\pi b^2 }$ with q a positive number of Coulombs. So the field jumps discontinuously from 0 to ${1 \over 4\pi \epsilon_0}\,{ -2q\ \over b^2 }$

From b to c, so b < r < c you have ${1 \over 4\pi \epsilon_0}\,{ -2q\ \over r^2 }$

At c there is another discontinuity. Approaching from r < c the value is as above, with the limit value ${1 \over 4\pi \epsilon_0}\,{ -2q\ \over c^2 }$.

In fact, the surface charge density at c is such that the jump at c ends up with E = 0 for r>c.
This is simply required from the fact that there is no E-field inside a conductor.
So this surface charge at c is induced by the charge on the inner sphere.
​

for c < r < d (d the radius of the outer sphere) the field is 0.

At d there is yet another discontinuity. Approaching from r < d the value is 0, and the jump is from 0 to $\sigma\over \epsilon_0$. Here $\sigma ={ +2q\ \over 4\pi d^2 }$ with q a positive number of Coulombs. So the field jumps discontinuously from 0 to ${1 \over 4\pi \epsilon_0}\,{ +2q\ \over d^2 }$

Finally, for d < r the field is like the field from a +2q point charge at the origin: ${1 \over 4\pi \epsilon_0}\,{ +2q\ \over r^2 }$.
with the limit value ${1 \over 4\pi \epsilon_0}\,{ -2q\ \over d^2 }$ for $r \downarrow d$.

No further comments.

---

​

 

[gracy]

No further comments.

You are locking this thread!(Although it's a mentor's job.)

 

[BvU]

No, I can't do that (lock a thread). And there is no reason to. But I urge you to read things through carefully and check if you have understood everything. Some digestion before firing of a bunch of "why ?"s and "right?"s. Did the considerable amount of assistance you received at least help you improve your mastership over this exercise ?

(And I appreciate your effort to pick up some TeX know-how!).

 

[gracy]

No, I can't do that (lock a thread)

I know it's a mentor's job.

 

[gracy]

And I appreciate your effort to pick up some TeX know-how!

You only taught me some.

 

[gracy]

I want to sum up this long thread in a post so that if someone visits this thread he/she would not have to face problems in keeping track
My OP contained two questions.These two question had been answered in post#7 and #4 respectively

1)why charge +4q is distributed /splitted in such a manner i.e +2q &+2q and then why one +2q is put in the circle of inner radius c and one +2q is placed in the circle of inner radius of d.

Such charge redistribution for the larger sphere is required to satisfy the requirement that the electric field inside a conductor must be zero. In order to have E=0E=0 for c<r<dc

2)what is the difference between the positions p3 and p4 because my textbook states at P3 electric field would have some magnitude but at P4 electric field would be zero because it lies inside the conductor then why P3 is not considered to be inside the conductor?

the two points are at different locations. The electric field at P3P3 is due to the opposing charges on the inside of the outer conductor and outside of the inner conductor. There must be an electric field at P3P3 as a result, whereas at P4P4 the electric field should be zero because you are on the conductor itself. P3P3 is not inside a conductor because it is outside the inner conductor which has a charge on its surface and P3P3 is inside the outer conductor which also has a charge on its inner surface.

Then as the problem progressed I included a graph in post #19
Electric field changes at the surface of a conductor .Just inside the conductor it is $\frac{σ}{ε0}$.In fact,the field gradually decreases from $\frac{σ}{ε0}$ to zero in a small thickness of about 4 to 5 atomic layers at the surface.
That's why graph shows two values of electric field at surfaces i.e at r=b,r=d ,r=c
Then I faced difficulty to comprehend why the value of electric field at r=c is negative as sigma is positive 2q?And it should nullify -2q and net charg enclosed should be zero.
I was wrong to think that gaussian surface at r=c includes /contains charge +2q .
Because charge enclosed by gaussian surface for r=c is -2q,.Because we are moving outward from the center of the sphere, and at r=c, we are yet to enclose the charge on the inner surface at r=c.
As+2q is not enclosed by it it is not going to nullify or cancel -2q .Hence net charge enclosed would not be zero rather -2q.

That's it!Thanks @ehild,@blue_leaf77 @Doc Al ,@BvU ,@Zondrina @haruspex for being patient and for giving helpful answers.You guys are amazing!