If we can shield a charge from electrical forces by putting it inside a hollow conductor. Can we shield a body from gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
9447
Electromagnetic waves can be shielded by a perfect conductor. What about gravitational fields or waves?
119763
Say we have a hollow conducting sphere (with some finite thickness).  If this object has an excess charge amounting to +Q coulomb, and there is no extra electric field in the surroundings (due to other charges), how will the charge be distributed?  Intuitively it seems that the charge will be symmetrically distributed..(all other cases seem too ugly), but in this case how is charge inside the conducting surface zero? Does this have something to do with solid angle and the fact that two cones with vertex at a point in the conductor will have a special relation of the (charge/distance^2) factors? (more charge is distance is more...)  If this is the case can someone please prove that the effect of the two corresponding cones will cancel each other out?
34551
Can we use  to calculate the force between two charges which are not at rest? If not, what formula should be used to calculate the force? I searched it, but I couldn't find a clear answer.
33024
I read this in my textbook: A charged particle or object is not affected by its own electric field.  Since I find this completely unintuitive and my mind is yelling "wrong! wrong! how could a particle even distinguish between its own field and the external fields?" I would really like to hear an explanation about why this is true, if it is.  (A corollary question that comes to my mind after thinking about this is... What about gravity and mass considering the analogous situation? And the other two forces?)
90824
When determining the gravitational attraction between 2 solid bodies, we can simplify computations by taking their masses to be concentrated at their respective centres of mass. However, had they been electrically charged bodies and we needed to compute electrostatic attraction, there is no "centre of charge" notion available to us. ( We would evaluate a 2-dimensional integral, or equivalently, apply Gauss's law. )  Is there any simple reason for the absence of such a centre of charge? Frequently, I've got the answer: "charge just doesn't work that way".
84812
Consider your typical  scenario.  You are standing on the inner side of the sphere.  You look up and see the shining sun far away.    My question (or thought experiment rather) is: Completely negating the gravitational forces, mass, etc. of the sun, if you could be placed anywhere within the radius of the Sphere, is there a point within the Sphere where you would float freely? The center seems the most obvious answer, but is it the only one? Keep in mind: standing at any point on the ground you have a "small" amount of mass beneath your feet, and FAR much more mass "above" you. Does this mass pull you more than the ground beneath your feet?
86341
I was having a conversation with my father and father-in-law, both of whom are in electric related work, and we came to a point where none of us knew how to proceed.  I was under the impression that electricity travels on the surface while they thought it traveled through the interior.  I said that traveling over the surface would make the fact that they regularly use stranded wire instead of a single large wire to transport electricity make sense.    If anyone could please explain this for some non-physics but electricly incline people, I would be very appreciated.
42515
The details are hazy as I recollect this from my nipper days.Many, many years ago.  “Persons were charged with stealing electricity by placing a large coil, somehow, near high voltage AC transmission lines.”  Is this feasible or was it perhaps an April fool’s article that I swallowed hook line and sinker in my youthful naivety.  There is a modern reason for this question.   Due to country wide cable and transmission tower spar actual theft. We are looking at placing modules with a gyroscope and Wifi module to form a mesh that will report if a tower is being “attacked”. Of course I will still have to test if the electronic gyroscope is sensitive enough for this. But one of the problems is going to be, how to power the unit? Solar will require an extra solar panel and battery. If I could tap into the radiated energy from the actual transmission line, great. Many modern SOCs don’t require much power and can sleep most of the time.  To clarify our needs. We are looking at building a device to sense if parts of the metal tower are being removed by scrap metal thieves. If too many parts are removed from the tower. It will buckle and earth the line. This is still at the feasibility stage.
22298
I just began studying electrostatics in university, and I didn't understand completely why the electric potential due to a conducting sphere is  $$   V(\vec{r})=\begin{cases}     \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}, &amp; \text{if $r \le R$}.\\ \\     \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}, &amp; \text{if $r \gt R$}.   \end{cases} $$  Where Q is the total charge and R is the radius of the sphere (the sphere is located at the origin).  I only understand the second part of this equation (when $r &gt; R$).  I know Gauss Law. know the charges go to the surface. I know the electric field strictly inside it must be zero. (I also know the electric field is not defined for a point that lies exactly in the surface). And I know $\vec{E} = -\nabla{V}$.  Therefore, I know the electric potiential inside the sphere must be constant. Let $C$ be this constant.  It seems that  $$ C = \lim_{r \to R^+} V(r) = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R} $$  But why is this true? My textbook says: because the electric potential must be a continuous function. But why? I am hoping for a non-experimental reason.  Please be precise when mentioning $r&lt;R$ or $r\le R$. Those are different and I get easily confused when people misuse those.
29836
Let us assume the universe filled with positive charge. About a particular point, all the positive charged particles will be symmetrical. Now consider a sphere of radius $r &lt; \infty$ and apply Gauss law. According to Gauss law the electric flux through the sphere will be some finite value as it contains some positive charge. But by symmetry the electric field through the sphere and hence electric flux will be zero. This is a contradiction.
97164
I'm currently reviewing electricity for the AP Physics C exam, and came across a problem or general topic I don't really understand that has to do with Gauss's law.  The general form of Gauss's law we learn in AP Physics C is $$\int E \;\mathrm{dA} = q_{enclosed}/ε_0$$  So when the enclosed charge is 0, as in the case of a hollow spherical shell with uniform charge distribution, it follows that the electric field is zero in the sphere.  However, when the charge distribution becomes non-uniform, the enclosed charge is still zero, implying that the electric field is still zero. However, if all the charge on the sphere is moved to one side, there should be an electric field directed from one hemisphere to the other. I'm wondering how this application of Gauss's law here is wrong and why.  When writing up this question I found this link:  where someone answered a similar question. However, I still don't understand why if flux is zero it does not mean the electric field is zero. Please keep in mind that I only have taken Calculus up to AP Calculus BC, which I think roughly corresponds to Calculus II.
1475
It is quite easy to derive the gravitational field intensity at a point within a hollow sphere. However, the result is quite surprising. The field intensity at any point within a hollow sphere is zero.   What exactly is the reason behind this? Except for, of course, the mathematics behind it. Is there any logic why the field intensity should be zero within a sphere? For example, it is logical to say that the field intensity would be zero at the center, as all the intensities cancel out. However, this cannot be the case for any point within the sphere.
51551
Since Newtonian gravity is analogous to electrostatics shouldn't there be something called negative mass? Also, a moving charge generates electric field, but why doesn't a moving mass generate some other field?
78539
Why is there no electric field in the hollow space of the dees of a cyclotron though it is open at one face? Is electrostatic shielding applicable for space(hollow space) inside open conductors(like dees) also?
33184
Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? I have completely no idea and I am inquiring about this as it is an interesting question that popped in my head while doing physics homework.
82087
What are the ways electrostatic charged objects leak charge in humid conditions?  Can airborne particles pick up charge by contact, then be repelled hence removing charge? If so would it be a significant factor?
48903
Suppose an electron outside a wire is moving with the same velocity as the electrons inside, then due to length contraction, there would be excess positive charges per unit area causing attraction of the election towards the wire, and the positive charges moving in this frame of reference with the same velocity of the electrons but in the opposite direction will create a magnetic field that wouldn't affect the electron that is not moving in this frame of reference. Now, if a grounded shield is placed around this wire, the electric field would not penetrate it and therefore there should be no attraction but then why is the electron attracted to the wire in spite of this shield?