6
votes
Accepted
How to solve the Gauss' law?
You can have a non-zero potential -- as long as it is constant in space. This will always generate a zero electric field, since in this case $\mathbf{E} = -\nabla\phi = 0$, which is indeed expected ...
- 306
3
votes
Why are fields described as force divided by mass or charge?
Why are fields described as force divided by mass or charge?
Because they follow from the classical universal law of gravitation and Coulomb's law.
The force that each of two masses or charges ...
- 56.1k
2
votes
How to solve the Gauss' law?
Ive been unfortunately trying to apply simple boundary conditions on your equation with no luck, because it's wrong. so First of all, let me correct you:
Laplaces equation for $r-$dependance is:
$$\...
- 4,386
2
votes
Accepted
Electric field along axis of polarized cylinder
Your first argument makes sense to me, except one point: the surface bound charge density is $P(r)$ on the top and $-P(r)$ at the bottom.
Regarding whether the displacement field $\vec{D}$ should be ...
- 448
1
vote
Accepted
The second uniqueness theorem in electrostatics
Thanks for posting this! I myself had a doubt that I couldn't resolve until I saw the theorem put this way.
Still, what you said needs some important corrections/additions..
It actually proves (not '...
- 26
1
vote
Why are fields described as force divided by mass or charge?
The definition of field, is there to tell us about the effects of the field on an object of unity value. most force fields have the parameter of the object they effect as a multiplier, hence when you ...
- 35
1
vote
Why the charge in each capacitors plate are equal in magnitude in series combination?
Suppose you have two capacitors in series like so:
————————||——————||————————
C1 C2
Consider only the central piece:
...
rob♦
- 71.4k
1
vote
How to solve the Gauss' law?
I don't quite understand this because there is not charge density, hence, no electric field.
This is wrong. There can be a non-zero electric field $\vec E(\vec r)$ at a point $\vec r$ even when the ...
- 7,614
1
vote
Will the potential energy is same in both the cases?
Short answer: Not really.
The answer is slightly different when talking about point charges vs distributons.
Given I have some charge distribution $\rho_{1}$ and some other charge distribution $\rho_{...
- 4,386
1
vote
Will the potential energy is same in both the cases?
It will be the same only if you ignore the electric field of the dQ's that you moved there first, that is, you only consider the electric field of the original charge Q at the origin. Otherwise you ...
- 144
1
vote
Accepted
I can't seem to figure out a way to compute a gradient without reference coordinates
We can use the identity
$$\nabla(A\cdot B) = A \times (\nabla \times B) + B \times (\nabla \times A) + (A\cdot \nabla)B + (B\cdot \nabla) A$$
So,
$$\nabla(-p_1 \cdot E_2) = \nabla[p_1 \cdot (\nabla V)]...
- 1,065
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