Potential and Electric Field in a Sphere ( Hollow or Solid) ??
345042
Electric field:\n\nE=(1/4*pi*epsilon0)*(q/r^2)--> [spherical shell at r>Radius]\n\nE=0 --> [spherical shell at r<Radius]\n\nE=(1/4*pi*epsilon0)*(q/r^2) --> [spherical distriburion at r<=Radius]\n\nor E=(q/4*pi*e0*R^3)*r [uniform charge]\n\n-----------------------------------------------------\nPotential field:\n\nit doesn't have a formula but u can find it with this integral:\n\nv=(1/4*pi*e0)S(dq/r)-> S is integral!
935341
1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
343981
In two charged plates, other than between the plates (and the edge effects), are there any field line? \nYES! other field lines are between the outer sides of the plates(depending on the charges of the plates and the convention used for the direction of the flow). To prove this, just connect the outer sides of the plates by a wire through an amper meter and you will see the meter moving.\n\nWhy is the electric field constant everywhere in the space between parallel plates? \nThey are not unless the area of both plates is inifinte and that the two plates are parallel.  In this case, the 1) direction of the field is the same every where since by definition the plates are parallel and 2) the absolute value is constant.  If not, there will be a surface current and the current would make the entire electric field (its magnitude) to be constant over the entire surface.
1044842
simply round... or shall i say spherical objects contains the most volume in the least surface area possible, in other words, the surface area of a cube is larger than of a sphere if they have the same volume and so on and so on, for an example in the case of an air bubble and a water drop, the surface tension effect tends to decrease the surface area thus creating a sphere, same thing with planets but here, gravity also has a roll to contain as much volume as close as possible and as you know planets were created from molten rock so the surface tension also played the same roll of decreasing the surface area here too...
544480
36pi  \nThe length of the whole is measured down the side of the \nhole. So the bigger the sphere, the greater the diameter of the hole. No matter how big the sphere is, the remaining\nmaterial is the same. So assume the hole is infinitesimal\nin diameter. The remaining material is all the material\n which is a sphere with diameter 6 and radius 3 inches.\nSo V=4pi(3)^3/3=4pi3^2=36pi
674108
The poles of any magnetized item will be at the extreme opposite points...in other words the positive will be as far as it can get from the negative.  In the case of a sphere this would be "top" and "bottom" not core and perimeter.  (Think of the Earth's magnetic poles!).
1047892
OK, let p be pi, OK?\n\nThe volume of the sphere is 4*p*r^3/3...\n\nNOW, the first derivative of that, with respect to the radius, is:\n\n4*p*r^2 dr\n\nSO, you can approximate the answer by:\n\n(4*p*3^2)*0.01\n\nThat'd give you 0.36*pi cubic inches, which would be b
174332
The surface area of a sphere is A=4*pi*r^2 where r is the radius of the sphere. The *volume* is (4/3)*pi*r^3.
238505
I'm not sure to what extend should I answer this. But just what I have in mind and I think might be related. \n-The molecules within the ball are closely packed together (they interact with one another) ; and this what gives solid their firm/fixed shape.\n-The energy of the molecules is low compared to energy of molecules of liquid or gas.The molecules basically stay fixed on their position and vibrate with minimum amplitude.\n\nThat's mostly about physics for ball. I'll add more if I remember more. Cheers
809750
Ratio of what?\nSurface area?\nVolume?\n\nI will go with volume.\n\nassume radius of sphere is = 1\n\nsphere = 4/3pi\ncylinder = 2pi\ncone = 2/3 pi\n\neat the pi's (get rid of them) and multiply what's left by 1.5\n\nSphere:Cylinder:Cone = 2:3:1\n\nI hope that helps
1137727
I like the joke!  To play along, I define:\nr = metric r  and \nd = metric d\n\nV = (pi/3)*(d^2)*(3r - d)\n\nThis will work for the tank being any fraction of a sphere, not just 1/2
42550
Potential, kinetic, chemical, electrical, and thermal.
788768
Just the usual stuff.  Use an electrolyte (H2SO4/HNO3), a couple of electrodes, a reduction table of half cell potentials, and some electricity.
882023
The first is a sphere of radius a centered at (0,0,a) and the second is a sphere of radius a/2 centered at (0,0,a/2). Both spheres are above the xy-plane and are tangent to each other at the origin. The smaller sphere is contained inside the larger.\n\nIn spherical coordinates, the equations reduce to\nrho=2a*cos(phi) and\nrho=a*cos(phi).\n\nHence the limits are\nrho: from a*cos(phi) to 2a*cos(phi)\nphi: from 0 to pi/2\ntheta: from 0 to 2pi.\n\nI am using the mathematician's convention where phi is the angle down from the z-axis and theta is the angle from the xz-plane. Physicists use the reverse convention, so be careful!\n\nDon't forget the rho^2 sin(phi) !
990247
A sphere.  It has an inside and an outside.
1013776
V of sphere = (4/3) * pi * r^3\n\n= (4/3) * pi * (2 sqrt 3) ^ 3\n\n= (4/3) * pi * (2^3) * (sqrt 3)^3\n\n=(4/3) * pi * (8) * 3 * sqrt 3\n\n=(32 sqrt 3) * pi   cubic units
318297
The derivative of the potential function is the gravitational force.\n\nPotential(At distance r) = G*Mass/d\nForce = Derivative(Potential)\nForce(At distance r) = G*Mass/d^2\n\nWhere G is the gravitational constant and r is the distance from the center of the sphere.\n\nSo when you are considering 2 bodies of mass M & m we get the formula (GmM)/r^2.