|
|
|
|
|

| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jun 2008 12:37:22 IST
|
|
|
what can be the definition of electric potential?
how can we derive the formula for electric potential?
|
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jun 2008 13:01:43 IST
|
|
|

|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
electric potential at a point can be defined / calculated as / by calculating the amount of work done by external agent against the electrostatic force in carrying a unit +ve charge from very large distance(infinity) to tht point
|
Beat others otherwise they will beat u |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jun 2008 15:49:32 IST
|
|
|
DEFINITION Electric potential at a point can be defined as the amount of work done to move a vanishingly small test charge from infinity to that point per unit charge of it.
DERIVATION T o find the electric potential at P at a distance r from the pt charge q. Let A be an arbitary point on the path joining infinity to the pt P where a vanishingly small test charge dq is kept. Distance of A from q is x and eleectric field at A be E where E= q/(4piEox^2)
To move the test charge towards q by AB = dx small work done dW= F.dx = dqE.dx = dqEdxcos180 = -dqEdx [integration sign=int] [infinity=@] Total work done W= - int[ (dqE) dx W/dq = -int Edx = -int{@ to r} q/(4piEox^2) dx = -q/(4piEo) int{@to r} x^-2dx = -q/(4piEo)[ -1/r + 1/@] W/dq = q/(4piEor)
|
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 2 Jun 2008 16:15:04 IST
|
|
|
payal is absolutely correct,,,
|
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
if i helped u plzzzzz rate me,,,,,,, |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|