what is the electric potential, V(r), due to an infinitely long line of charge?
452540
At r = infinity, E(infinity) = 0.\n\nAt r = R, E(R) = constant/R.\n\nThe change is\n\nE(infinity) - E(R) = 0 - constant/R =  - constant/R.\n\nDon't drop the negative sign.
566490
the rate of change of charge density is constant along the cylinder. So if we consider two nearby points along the cylinder the charge difference between the two points would be the same aslong as the distance between them was the same(no matter where in the cylinder those two points were).\n\nSince the electric field is dependent on this difference then it will be the same all along the cylinder. (This assumes that all the other charges can be ignored because they either cancel with other charges or they are very small  (electric field is proportional to 1/r^2))
972428
C=80uF\nV=Voc1.5 volts\nI=Isc=5.0 amps (assuming short circuit current)\n\nSo thevnin's theorem implies the \nRth = Voc/Isc = 1.5/5 = 0.3ohms\n\nSo the circuit is an RL circuit,\n\nTime constant T = RC = 0.3*80uF = 24us\n\nNow you can use that formula for charging
127709
Coulombs law.\nWhen the distance is made 3 times longer (for instance),\nthe force between the charges becomes 3^2=9 times smaller.
355899
The electric field at a point in space is equal to force per unit charge exerted on a positive test charge placed at that point.\n\nConsider a point charge +q located at point. The charge +q sets up electric field in the space surrounding\nit. If a small positive test charge +Q placed at point P experiences a force F.Then electric field ,E=F/Q\nMind the following:\n(i) Electric field (force/charge) is a vector quantity, i.e., it has magnitude as well as direction.\n(ii) The SI unit of electric field is newtons per coulomb (N/C). As shown later, it can also be expressed\nas volts per metre (V/m).\n(iii)The other terminology used for electric field is electric field intensity or electric field strength.
32356
Consider a system of charges. The charges of the system exert electric forces on each other.If the position of one or more charge is changed, work may be done by these electric forces.We define change in electric potential energy of the system as negative of the work done by the electric forces as configuration of the system changes. Potential energy of the charge particle in an electric field depends on the magnitude of the charge.However, potential energy per unit charge has a unique value at any point in an electric field.For eg. We place a positive charge x Coulomb at a point in an electric field where particle has an electric potential energy of y Joule. Then potential energy per unit charge is =x/y J/C(Using unitary method)\nIf we replace our charge by another charge of magnitude 2x Coulomb.We would find that the second charge has a potential energy of 2y Joule.Potential energy per unit charge would be same.Potential energy per unit charge is independent of the charge we are using.\nThe potential energy per unit charge at a single point in an electric field is called the Electric Potential.Now the difference between the Electric Potential between two points is called Potential Difference between those two points.\nPotential difference is directly proportional to current.(Ohm’s Law)\n\nFor further queries e-mail me:  trust_amplifier@yahoo.com
313019
This is a simple application of Coulomb's Law.\nThe Electric Field is simply given by the charge on the nucleus (26e) divided by 4*Pi*EpsilonNought*distance^2\nWhere e is 1.6*10^-19 Coulombs
603730
The following refers to Coulomb force between two charges in a single uniform medium, not the two-medium situation described in  the question.\nAnswer 1 is helpful but the lack of exponent signs makes it misleading. The force is inversely proportional to the square of the distance, so the simplified formula is \nF = k * q1 * q2 / r^2 \nk is a constant = 8.99*10^9 N m^2 / C^2 (C is charge in Coulombs). Thus if q1 = q2 = 1 Coulomb, and q1 is separated from q2 by r = 1 meter, the force is 8.99*10^9 N or 2.01 billion pounds! \nThe above simplified formula is a contraction of the full formula (see ref.),\nF = q1 * q2 / (4 * pi * epsilon * r^2)\nin which k is replaced by 1/(4 * pi * epsilon), where epsilon is the product of epsilon0, the electrical permittivity of the vacuum (= 8.85 * 10^-12 c^2/(N m^2)), and kappa, the dielectric constant which = 1 for vacuum and is assumed in the simplified formula.\nFor two segments of distance as you specify, first solve for F1 and F2, each as a function of a particular distance (r or d - r) and dielectric constant. Then my inclination would be to try the sum-of-inverse square root solution which works for a similar situation, gravitational force F at distance R, computed from F1 and F2 at partial distances R1 and R2, where R = R1 + R2, \n1/SQRT(F) = 1/SQRT(F1) + 1/SQRT(F2)\nSorry I can't be more definite than that.
31718
for an ohmic material:\n\nP=VI\nV=IR\nso\nP=I^2 * R\n\ncheers!
268827
That's the definition of equipotential lines, basically.\n\nIf you move perpendicular to the electric field, you do no work (so potential doesn't change).
792406
let the initial charge on the capasitor be q. Then for charging the equation is given by:\nQ=q(e)^(-t/T);  where e is base of natural log. 't' is the time and 'T' is time constant which is probably 1/(RC)\n\nfor discharging\nQ=q(1-(e)^(-t/T));\n\nnow the graph can be easily plotted by taking Q on 'y' axis and 'q' on x-axis.
2470
I'm assuming this is an electric dipole.\n\nSince a dipole can be modeled as a pair of opposite charges +_q separated by a small displacement, you can use the method of images and assume the field produced by the conductor is the same as the original dipole reflected across the plane with the signs of the charges reversed. That means the image dipole is in the same orientation as the original dipole and is at a distance r = 2I from it. The force is attractive (towards the plane). The energy between two dipoles having the same moment p parallel to their separation vector is:\n\nW = -2p^2/r^3\n\nwhere r is the separation distance. The force between two such real dipoles would be (i.e. this is the force on either one):\n\nF = dW/dr = -6p^2/r^4\n\nThis should be the same force experienced by the one dipole in the field produced by the conducting plane. Since r = 2I, this is:\n\nF = (3/8)p^2/I^4
27954
You would need to know what voltage the capacitor was charge to. Volts * capacitance = Coulombs
438763
Potential difference (V) across the capacitor is directly in proportion to the charge (Q) in coulombs, calculated by Q =CV   Remember that C must be in Farads, so 1 microfarad is 1x10^-6 not 1, etc.\nSo values up your pd or Volts axis on the graph can be multiplied by capacitance to yield charge (Q).\nThe change in charge Q through the time period in question divided by the length of the time period (in seconds, not milliseconds etc) gives the average rate of movement of charge in Coulombs per Second (= Amps) over that period.\n(Also, the gradient at any point of the trace gives the current at any instant, but that is not what you asked.)
282905
a) E=v/l ---> E = (12/3)*1000=4000 (Volt/meter)\n\nb) v ~ 1/r ---> v2=12*3/5=7.2 Volt\n   E=(7.2/5)*1000=1440 (Volt/meter)\n\nc) w=int(f.dx)=int(Eqdx)=int(vqdx/x)=int(36qdx/x^2)\n\n    = 36q*(-1/5+1/3)=72q/15\n\nq=4*pi*e*l*v/2=2*3.14*8.854*10^(-12)*0.003*12 = 200*10^(-12)\n\n\n---> w=960*10^(-12) J
738850
"A potential gradient (E) is the local space rate of change of the potential.\n\nIn electrostatics then, it is the local space rate of change of the electric potential:\n\n    E = - delta V / delta s\n\nUnits are volts per meter (V/m). The electric field is the same as the potential gradient but with opposite sign."\n\nFrom Wikipedia,\nhttp://en.wikipedia.org/wiki/Potential_gradient\n\nHope this helps!
331292
See the other postings of this question.  The equation is\n\nhttp://members.aol.com/logan20/elecspon.html