what is da equation to calculate electric field?
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.
982407
dy/dx=cosx*e^y*e^sinx\ndy/e^y= cosx*e^sinx*dx\ne^-ydy= cosx*e^sinx*dx\nintegrate\ne^-y/(-1)=int. cosx*e^sinx*dx\nput sinx=t\ncosxdx=dt\n-e^-y=int.e^tdt\n        =e^t+c\n        =e^sinx+c\nput x=0,y=0\n-1=1+c\nc=-2\n-e^-y=e^sinx-2
753535
Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
276990
read this page\nhttp://en.wikipedia.org/wiki/Field_(physics)
373192
f=m*a=E*q\na=Eq/m\n\nv^2=v0^2+2ax\n\nv=sqrt(v0^2+2ax)=sqrt(v0^2+2Eqx/m)\n\nv0=2.74e4\nE=2.58e3\nq=1.6e-19\nx=.00185\nm=1.67e-27
560737
En=-(13.6)/n^2eV\nE1=-(13.6)eV\nE3=-(13.6)/9eV\ndelta E=E3-E1=hv\nhv=12eV\ntherefore,v=(12*1.6*10^(-19))/(6.626*10^-34)\nv=2.8976*10^15hz
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.
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
810779
Let the two +Q charges be placed left to right with the distance r separating them.  Let a third charge of unknown charge q be placed a distance x to the right of the first charge.  Now, look at each of the +Q charges and sum the electric force acting on them.  For example, the left hand +Q charge experiences a repellent force from the +Q on the right, pushing it towards the left.  To balance this out, there must be an attractive force pulling it to the right, so the q charge is negative.  However, this is just an observation.  In order to maintain the integrity of the algebra, we should continue to assume that q is positive, but recall the negative result llater when we check our answer..  The proportionality factor for electric field strength can be ignored, because it will cancel out, so you can treat Q/r^2 as the value under consideration.  In this case, we need Q/r^2 + q/x^2 = 0.  However, because the +Q charges are equal in magnitude, I will make the assumption that the distance x = r/2, so Q/r^2 = -4q/r^2 ==> Q = -4q ==> q = -Q/4.  Perform the same force balance on the right hand +Q charge to verify.\n\nIf the initial charges had been unequal, x would have remained a variable, and the complementary distance (r - x) would have been used for the second force balance in order to form a system of equations.
202486
E(sub l) = L(di/dt). \ni is the varying current through the circuit during transient period or in AC circuit.\nif it's DC, E(sub l) will be equal to 0 after some time..
211994
V = Er = 3E6 x 0,30 = 9E5 volt = 900 kV
368244
V (Voltage) = I (Current) x R (Resistance)\n\nOr (although you probably don't need it):\n\nJ (current density) = sigma (conductivity) x E (electric field)\n\nI always use wikipedia to look up stuff like this, it's very helpful for all kinds of info.
892848
One of the most frequently used is E=IR    ;E =voltage    I=current   R= resistance. U can change this every way depending on the unknown.\n    Good luck
911239
In standard conditions \ndE = -0.0592 V/n log [Fe 2+]/[Cd2+] = - 0.0592 V/2 log 0.19/1.2 = 0.024 V\n\nYou can take a look here for explanations and examples:\n  http://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html
948228
D =M/V\n\nM=DV\n\nV=M/D
786148
Electronic engineering - circuit analysis:\n\nSolving a circuit for the current and voltage according to Kirchhoff's Laws and these device equations:\nResistor   : v = R * i\nCapacitor: i = C * dv/dt\nInductor   : v = L * di/dt\n\nAnything complicated you generally solve by using the Laplace transform, in which case the above equations simplify to\nfor V,I = Laplace transforms of v,i:\nResistor   : V = RI\nCapacitor: I = sCV\nInductor   : V = sL\n\nThen you solve the system of equations for V,I and apply a reverse Laplace transform.
566230
Set up the equation\n\nP = dRT\n\nAssuming P to be constant,\n\ndT = d'T'\n\nor d' = 0.76*273/298 = 0.696 g/l.