1someone solve it man
the charge q is being brought in from infinity
dWfield = Fdr cos180
→dWext = Fdr
du=dwext= Fdr
r r
∫du=∫Fdr
∞ ∞
U =- kQq/r (Assuming potential energy at infinity to be 0)
Here i am bringing in a positive charge against the field hence conceptually the potential energy must be +ve but mathematically it is coming negative.Why?
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9 Answers
23ur r vector and dr vector are in opposite directions .....
so there is something wrong
23i m not sure if the following is correct but abhi dimaag me yahi a raha hai
→ →
F . dr = |F| |dr| cos 180
= (F)(-dr)(-1) ( since magnitude of dr = - dr since dr is negative )
=Fdr
this quantity (Fdr) i.e work done by field is negative since dr is negative as per ur dr
so work done by ext will be -Fdr and u wil hav to integrate - Fdr and not Fdr
1work must be done by an external agent to rotate the dipole through a given angle which is opposite to the direction of rotation caused by the torque due to field.
this work gets stored as potential energy provided kinetic energy is kept constant
dWext=td\theta
this work gets stored as potential energy
U(\theta)-U(90)=∫pEsin\thetad\theta (from 90 to \theta)
→U(\theta)= -pEcos\theta
again in this case...i have done work against the field and still come out with negative potential energy....??????
23The reason is that u hav taken PE = 0 at \theta = 90
see for example
in gravitation ,
suppose there was only earth in the space and nothing else
logically speaking PE shud be maximum at infinity , bcz if i leave a body at infinity, then it will crash on the earth's surface with greatest KE .
releasing the body from any other finite distance wont hav this much KE on reaching the earth
SO logically PE shud be max at infinity
But in gravitation , we choose PE = 0 at infinity
so wat shud be the PE of those points wich are at a distance less than infinity ??? of course their PE shud be less than that at infinity
so
PEat a finite point < PE at infinity
hence PE at a finite point < 0
as we assumed PE = 0 at infinity
if we had chosen PE =0 on the earth's surface then we would get PE > 0 at all points above the earth
its just for simplicity that we choose PE = 0 at infinity
and yes wen we say that PE of a body at a point is something
it is actually the change in PE of the body when we bring it from the reference point to that point
see PE of a body at a point will change if we assume PE = 0 at different unidentical points
but difference in PE between 2 points will never change , even if u change ur point of reference or point of PE = 0
the logic is same for PE of a dipole
if u had chosen PE = 0 at \theta = 0 , then ur PE will turn out +ve for a \theta > 0
1Electrostatic PE is taken as 0 at infinity.
when the charge is brought near another charge, work is done against the conservative field which is stored as PE.
Work done= - PE
work done by charge is +ve so PE is -ve.
23@ ishan , for ur first question , see #4
if u dont agree
then recall that |x| = - x for x < 0
ur dr < 0
so | dr| = - dr
and A . B = |A| |B| cos\theta