Theoretically (I use that loosely) and most commonly identified as the electrical potential. Look in a physics book next time, but the formula for is U = kQq/r. I'm not giving you what the variables stand for because I am a dick. Also because you need to try to learn something but RESEARCH, not stupid questions.
KTHX BAI.
Potential difference
potential difference is the amount of work done in moving a unit charge from one point to another point. potential difference be V ,work done be J and charge be C.V=J\C
Yes, that's what it means. No force would be required to keep a test point-charge moving along a line of zero potential in the direction toward that point, and there would be no force attracting it toward that point in the combined field. Of course that's physically impossible in the real world, probably because there's no such thing as a point charge. The smallest possible test-charge would still have some non-zero physical dimensions, and be made of atoms whose charge distribution inside it is non-uniform. So it could never stay exactly on the line, and any slight perturbation would require force to execute a mid-course correction and put it back on the zero-potential. Even if there is no continuous contour of zero potential available for the trip, if the test charge starts out and arrives at points of zero potential, then the work done along the way to push it against an occasional repelling force is exactly equal to the work done by an occasional attracting force, and they add up to zero for the trip.
It will be directed away from the positive charge. It will attract any other negative charge and repel any positive charge. Its magnitude is given by E= KQ/R where K = 9x 109 C2m-2N-1 Q is the charge producing field R is the point where electric field is to be calculated
consider value of charge =e electric field intensity =E(xyz) now acharge inelectric field experience a force of =eE(XYZ) NOW WORK done to move this particle of distance dl (small lenth) is =eE(xyz).dl (you have to take dot product) now to move this charge praticle from point (x1,y1,z1) To (x2,y2,z2)=intregration of [eE(xyz).dl](x1y1z1)to (x2y2z2)
Potential difference
The work to be done to bring a unit positive charge from infinity to a point in an electric field exists which is having a magnitude unity and direction opposite to the movement of the unit charge.
it is defind as the amount of work done to bring a unit positive charge from infinity to that point in the electric feild it is devoted by V .: electric potential = workdone/charge V=w/q si unit is v
No, work done in moving a charge from infinity to a given point does not involve any acceleration. Work is defined as the product of force and displacement, and in the case of moving a charge, the force is constant along the path. Since acceleration is the rate of change of velocity, and there is no change in velocity in this case, there is no acceleration involved.
Everything. A positive charged particle generates an electric field equivalent to the work done in bringing a unit positive charge from infinity to near that charge.
potential difference is the amount of work done in moving a unit charge from one point to another point. potential difference be V ,work done be J and charge be C.V=J\C
No one because infinity is not a number.
then our work is positive
Yes, that's what it means. No force would be required to keep a test point-charge moving along a line of zero potential in the direction toward that point, and there would be no force attracting it toward that point in the combined field. Of course that's physically impossible in the real world, probably because there's no such thing as a point charge. The smallest possible test-charge would still have some non-zero physical dimensions, and be made of atoms whose charge distribution inside it is non-uniform. So it could never stay exactly on the line, and any slight perturbation would require force to execute a mid-course correction and put it back on the zero-potential. Even if there is no continuous contour of zero potential available for the trip, if the test charge starts out and arrives at points of zero potential, then the work done along the way to push it against an occasional repelling force is exactly equal to the work done by an occasional attracting force, and they add up to zero for the trip.
zero along the direction of the field
Correct, no electrical work would be done. There may be some mechanical work done by the mechanism that moves the charge and this will probably be converted to heat.
It will be directed away from the positive charge. It will attract any other negative charge and repel any positive charge. Its magnitude is given by E= KQ/R where K = 9x 109 C2m-2N-1 Q is the charge producing field R is the point where electric field is to be calculated