googly from ATGS 15

Let ' s look at the behaviour of the phenomena caused by the charge " q " placed in front of the conducting plane . First , it will induce some charges on the plane itself . Then , there would arise a mutual attraction , due to which the charge " q " will gain some potential energy . But , if we try to move away the charge , then the charge distribution of the induced charges will change . In fact , the charge distribution on the plane at a distance " x " from the foot of the perpendicular drawn from the charge " q " when it is at a distance " d " from the plane could be derived as -

σ = - q d2 Ï€ ( d ² + x ² ) ^{3 / 2}

Hence , if the charge distribution changes , then there would arise a field , which obviously will become time - dependant . So , it no longer will satisfy the concept of potential , as " Potential " could only exist in case of time invariant fields .

yaar can neone plzz explain me image method with an example plzzzzzzzzzz????????????

Well here the concepts of potential field is still applicable. That's why the word "slowly" is stressed so that the field remains quasi static. The potential method would give the correct result provided you divided by 2. The reason for this factor of 2 becomes clear if one understands the basics of image method. The point is that we are considering two separate problems. One in which we have a point charge q at a distance d from a conducting plane. In the second case, we have actually two point charges. Only the halves of space (divided by the plane) in which the point charge q is present are electrically the same. So while considering the work by potential energy method, we must remove the contribution from the other half of the space where the negative charge is present.

The second process is correct . First of all , the charge placed will induce some charges on the plate , which in turn would create a field . But , the field caused by the induced charges is not a " Potential Field " . To make things sound more clear , it would be better if I say the integral of this field around a closed path is non - zero . That is why the concept of " Potential " fails here .

In other words , Work Done = Loss In Potential Energy

this relation will not be true in this case .

i think first method is wrong because the energy content of the system is spread on both sides of the plane but in our actual question and the energy will be above the plane and not below. thus , potential energy will be half.