how to derive the potential on the surface of a charged sphere ?
460932
Assume all the charges are concentrated at the center of the sphere.\n\nNow the problem is reduced to find the potential at a point which is at a distance equal to the radius of the sphere from the center of the sphere where all the charges are accumulated.
744501
C = Q/V = εA/d\n\nwhere ε is the permittivity of free space or air\n= 8.8541878176 × 10−^12 F/m (C2/Jm)\n\nQ = (εAV)/d = [(8.85×10^−12)(0.4)(200)]/(0.0005)\n= 1.416 x 10^-6 Coulomb\n\nis the charge on the upper plate.\n\nThe charge on the lower plate that is grounded is ZERO, because the charges have moved down into the ground.
913226
The equation was developed by Coulomb:\n"In physics, Coulomb's law is an inverse-square law indicating the magnitude and direction of electrostatic force that one stationary, electrically charged object of small dimensions (ideally, a point source) exerts on another. It is named after Charles-Augustin de Coulomb who used a torsion balance to establish it."\n\nsee Wikikpedia;\nhttp://en.wikipedia.org/wiki/Coloumb%27s_law\n\nF = Kc * (q1*q2)/r^2\n\nwhere,\nF = force magnitude\n\nq1 = charge on one body\n\nq2= charge on the other body\n\nr = distance between charges\n\nKc = coulomb force constant = 1/(4*pi*epsilon_0) ~ \n8.988×10^9 N m^2 C^-2 \n\nepsilon_0  ~8.854×10−12 C2 N-1 m-2 (also F m-1) is the permittivity of free space, also called electric constant\n\nYour question:\nF = 4*10^-5\nr = 3^10^-2\nq1 = q2 = q\n\n4*10^-5 =  8.988×10^9 * q^2/ 3*10^-2\n12*10^-7 = 8.988×10^9 * q^2\nq^2 = 12*10^-7 / 8.988×10^9 \nq^2 = 1.335 * 10^-16 C\nthen,\nq = 1.155 * 10^-4 C\n\nHope I'm correct, and hope this helps!
621034
Gravitational potential or electrical potential?
632601
isthegeometricaltheoryofgravitationwhichunifiesspecialrelativityandIsaacNewton'slawof universalgravitationwiththeinsightthatgravitationisnotduetoa forcebutratherisamanifestationofcurvedspaceandtimethis curvaturebeingproducedbythemass-energyand momentumcontentofthe spacetime
45249
Let S be a surface in three-dimensional space ( R^3 ) and let f be a function defined on this surface. To describe the surface, we will choose a parameterization of the surface by two variables which we shall call u and v. (For instance, if f is a sphere, u and v might be the latitude and longitude.) In terms of this parameterization, the function f can be expressed as a function of u and v. Then the integral of f with respect to the surface area is usually notated as \n\nSurface Integral S[f(u,v)d^2 A]\n\nJust as with the ordinary integral, this quantity may be understood as a limit of sums.1 To form an approximating sum, we subdivide the surface S into a miniscule pieces, multiply the area of each piece by the value of f at a point located inside that piece, sum over all the miniscule pieces into which we have subdivided the surface. The limiting value of these sums as the size of the pieces shrinks to zero is the integral with respect to surface area. \nTo compute this quantity, one may use the following formula to convert it to a double integral: \n\n\nIf the surface is described as the graph of a function , then we may also use the following formula:
728779
Sounds like youve got a nasty integral. Your going to have to integrate a polar vector in cartesian coordinates. Write the B field in cartesian and then integrate one side of the square.\n\nB.ds= miu (0) * I is true. I forget which theorem but I think its stokes theorem from vector calculus. It doesnt matter which shape you choose. The symmetry of the circle allows a simple integral because there is no dot product for the B.s since they are always parrallel. Choosing a square just makes your life horrible. If you correctly write the B field and then use the vector identity:\n\ncos theta = a.b/(|a||b|)\n\nto find the angle for the dot product you will get the same answer. It's just hard to set up the problem.\n\nIm sorry dude, good luck.
1064629
find the surface area first (thats easy). Then given that the emissivity of th sphere is .851, and that the rate of change will be the change of temp/time. Since there is no time involved directly, we need to have the Surface area x emissivity rate / the temperature difference (in relation to the time, which is given within the emissivity rate). This should give u the net rate, depending on the emissivity/time relationship. You do the working out, otherwise you won't learn it.
197297
This would be a function of vectors. Assuming you are reffering to sine waves the inverse of the sine gives you the angle. The total energy would be a function of Root Mean Square.
1175686
This can be solved in several different ways.  One of course is by calculus, by developing a formula for the surface area of a "slice" of the remainder of the sphere and integrating it about a rotated axis, or something similar.  Another is by digging for the rather obscure formula for the "volume of the endcaps of a sphere" and subtracting two of those and the volume of a cylinder from the volume of the original sphere (and discovering that many of the terms will cancel out nicely).\n\nThe easiest method, however, is via the use of simple logic.  Unfortunately, this method also requires one more "given" in order for it to be valid, though this given is not an unreasonable one:  There must be a single, real answer (or more specifically, a unique answer on the real number line).  Such an assumption would be folly in a practical (real-world) context, since without some form of proof, one should never immediately assume that this is the case.  Perhaps there are multiple answers, or there is no answer at all?  But as luck would have it, there actually is one answer, and it may be derived logically due to its unique existence.  (*thumbs nose at any philosophers who may be reading*)\n\nSolution:\n\nLet R be the radius of the cylinder drilled into the sphere.  If and only if there is a single, real answer, then that answer would still be the same no matter the value of R... even if R=0.  Thus, the answer is the volume of a sphere of radius 3 (diameter 6), or 36*pi cubic units.\n\n(Yes, the volume is always the same, whether the original sphere were the size of a basketball, or the size of a planet.)\n\nIt's a surprising answer, along the lines of the more familiar puzzle: "If the Earth were a perfect sphere and a band were placed snugly about its equator, and then one more meter were added to the length of the band and it were stretched so that all points along it were equidistant from the surface of the Earth, how far above the Earth would the band rise?"  Many are surprised to find that the size of the Earth plays absolutely no part in the equation, and that the "rise" is exactly the same if the sphere were basketball-sized or, umm... Earth-sized (with a circumference greater than 40,000,000 meters).  It rises exactly the radius of a circle with a 1-meter circumference: 1 / (2*pi) meters.
418584
The rate of increase of the volume dV/dt can be solved as dV/dr*dr/dt (chain rule)\n\nYou are given that dr/dt = .04 cm/s and you can derive dV/dr from the volume formula.  The derivative of r^n is n*r^(n-1) so the derivative of r^3 is 3r^2.  The derivative of the volume formula is 4/3*pi*3r^2, or 4pi*r^2.  Therefore the sphere is growing at 4pi*r^2 in volume per radius length.\n\n(a)\nif r = 10cm then the rate of increase of volume, dV/dt = dV/dr * dr/dt = 4*pi*(10cm)^2*(.04cm/s)\nDo the math and you should get an answer in cm³/s\n\n(b)\nGiven V=36, find r.  The area of the cross section = pi*r².  The rate of increase of that area is 2*pi*r*dr/dt.\n\n(c)\nIf dV/dt and dr/dt are the same number then dV/dr=1.  Find r so that dV/dr=1.
757387
(a) Assuming the base is a half-circle and that there are no other forces acting on the mass (friction, drag, etc), then the kinetic energy at the base is equal to the potential energy at the top. Note that the change in potential energy is not affected by the semi-circular path the mass follows, but rather the net change in height. This net change is simply the radius of the half-circle container:\n\nKE = mgR = (.71 g * 1 kg/1000 g)(9.8 m/s²)(25 cm * 1 m/100 cm)\n\nKE = .00174 J\n\n(b) The kinetic energy associated with rotation is given as:\n\nKER = (1/2) I w², where w is the rotational velocity. Note that since the sphere rolls without slipping, the translational velocity of its cener of mass is simply:\n\nv = w * r\n\nSubstituting this into the expression for the rotational kinetic energy:\n\nKER = (1/2) (I / r²) v²\n\nNote that the expression for the translational kinetic energy is:\n\nKET = (1/2) m v²\n\nThus, the total kinetic energy of the ball at the base of the hemisphere is;\n\nKE = KER + KET = (v²/2)( I/r² + m)\n\nThe fraction of the rotational kinetic energy over the total is:\n\nf = KER / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)\n\nf = ( I/r²)/( I/r² + m)\n\nFinally, the moment of inertia, I, of a sphere is:\n\nI = (2/5)mr²\n\nSubstituting this into the expression for f:\n\nf = (2/5)m/( (2/5)m + m)\n\nf = (2/5)/( (2/5) + 1)\n\nf = 2/7\n\n(c) At the base of the hemisphere, there are two forces acting on the ball: the force of gravity and the centripetal force. At the base, these forces are coincident (act in the same direction). Thus the normal force, which opposes these two, is simply:\n\nN = Fgrav + Fcent\n\nN = mg + mv²/R\n\nTo find v, recall that KET / KE is 2/7. Using this, we can manipulate the terms to get v:\n\n(1/2) (I / r) v² = (2/7)(mgR)\n\n(1/2) ((2/5)m) v² = (2/7)(mgR)\n\n(1/5) v² = (2/7)(gR)\n\nv = sqrt((10/7)(gR))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = 1.871 m/s\n\nFinally, the normal force is computed as:\n\nN = mg + mv²/R\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = .0169 Newtons\n\n\nGood Luck! :)
969797
Let the radius of the sphere be R.\nLet x be the radius of the cylinder.\nThen height of half of the cylinder= sqrt(R^2-x^2)\n\nTotal surface area of cylinder = 2*pi*x^2 +2*pi*h, so\nS=2*pi*x^2 +2*pi*2*(R^2-x^2)^.5  [Eq I]\ndS/dx=4*pi*x + 4*pi*.5*(-2x)(R^2-x^2)^-.5\n=4*pi*x - 4*pi*x/(R^2-x^2)^.5\n=4*pi*x(R^2-x^2)-4*pi*x =0\nR^2-x^2-1=0\nx^2=R^2-1\nx=+ or - sqrt (R2-1)\nPlug these values of x into [EQ I] to get answer.
205480
Surface area of a sphere is = 4 pi r^2\nThis means that surface area of a sphere is equal to 4 times the biggest circle in the sphere Or 4 times the area of the circle swept by its radius or 4 times the area of the base of its hemisphere.For example if the radius of a sphere is 1 unit then its surface area is = 4 pi square units.When it is divided by 4 it is equal to the area of the base its hemisphere.ie. area of the base of its hemisphere is = pi square units.
549577
The formula for surface area of a sphere is \nA = 4*pi(r^2)\n64 pi = 4*pi(r^2)\nwhich means r^2 = 16 or r = 4\nVolume of a sphere is \nV = (4/3)*pi (r^3)\nV = (4/3)*pi (4^3)\nV = (4/3)*pi (64)\nV = (256/3)*pi
418581
The earth is taken as the reference point of measuring potentials and the potential given to earth is 0(after all only potential differences make sense) . The earth has such an enormous radius, that its capacitance\nC=4*pi*e*R is very very large and hence the change in potential of the "sphere" is negligibly small as\n change in potential = charge added/capacitance of earth\nThus the earth remains "neutral"(zero potential assigned initially doesn't change) even after lightenings.
473554
You have to use integration by parts.\nLet u = ln(x) and dv = dx, then\ndu = (1/x)dx and v = x\nint(udv) = uv - int(vdu)\n= xln(x) - int(x(1/x)dx)\n= xln(x) - int(dx)\n= xln(x) - x + C, where C is a constant