11some one plz try
if charge is distributed uniformly in the x-y plane with charge density +(sigma) in 1st quad and -(sigma) in remainig quad then work done in miving a point charge q from (0,0,d) to (0,0,2d) is
A) -(sigma)qd/4e
B) (sigma)qd/4e
C) -(sigma)qd/2e
D) (sigma)qd/e
(e= epsilon not)
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7 Answers
1work is a state function
and as there is no electric field in the z plane and the body is displaced along the z axis hence the ans is 0
11the electric field for two OPPOSITELY chargsheet will canCel out.then the force is only due to two negatively charge sheet.
WORK DONE=F.dS
F=QE
E=SIGMA/2E0+SIGMA/2EO=SIGMA/E0
F=Q(SIGMA)/E0
dS=0-0+0-0+(2d-d)=d
w.d=q(sigma)d/E0
OPTION NO D IS THE CORRECT ANSWER
11sorry.the rigt solution is this
the total electric field due to the total charge sheet is ∂/2E0
(if we consider the total quadrant as a charge sheet)
then electric field due to one charge shhet is=∂/2EO*4=∂/8E0
THE ELECTRIC FIELD FOR TWO CHARGE SHEET IS CANCELLED OUT
THEN TOTAL ELECTRIC FIELD IS=-∂/8E0-∂/8E0=-∂/4E0
WORK DONE=F.dS
dS=(2d-d)k
f=(-∂/4E0)k*q
w=-∂qd/4E0
62See what will happen is that there will be electric field in x, y and z coordinates.
The component of the the z coordinate does not change if we remove some of the charges symmetrically... (I dont know if this makes perfect sense to each of you)
The two sheets will cancel each other (in the 1st and the 3rd)
We take the 2nd and the 4th quadrants.. The electric field will be σ/(2ε0)
The charge density is half of sigma (because only half the plate is charged)
So the electric filed will be -σ/(4ε0)
Now you have to find the work done which will be F.ds, Thus
A) -(sigma)qd/4e