IIT JEE , AIEEE Forum - Physics , Electricity

anshul_2009

Hi ,

Lets consider we have a point charge around which its elcetric field is prsesnt, so if we assume gaussian surface as sphere around it , such that a point chrge is present in the spherical ball so we use gauss theorem tofind the electric flux .

But suppose point chrge is not in the gaussian surface but it is outside and near , then what will be the flux. Plz reply as soon as possible

computer001

ur q does not make sense
u urself have stated tht to find the flux u r considerin a spherical surface arnd the chrg...then wat do u mean by the chrg is outside a gaussian surface??

anshul_2009

Suppose the box is present as shown in figure and a point chrge is also there .How to find the flux then?????

hash_include

flux is 0.

anshul_2009

ya but why an u explain me ?????????

computer001

anshul it depends on wat u want to find the flux thru..
acc to ur dig flux thru the cube is 0..
as chrg encl in the box is 0...
wateva field lines enterin the box will leave so flux=0

seriousstone89

NET FLUX = 0

guass law says net flux not flux

net flux is zero because the flux entering the suface is equal to the flux leaving the surface.

ananth_patri

net flux is zero as the Gauss law is only for q inside which gives the flux,,,,,,as the qinside is 0 the flux is also zero....

ayshwarya

according 2 gauss law flux can b calculated thru a given closed surface hence 4 d given quesn flux is 0

hope dis resolved ur quesn

rate if satisfied

ayshwarya

fr more clear information u can refer h.c verma-ii

SCIENTIST135

FLUX BY GAUSS LAW CAN BE FOUND ONLY WHEN A CHARGE IS PRESENT INSIDE A BODY ANS NO CHARGE IS INSIDE THE CUBE, FLUX IS ZERO. ALSO GAUSSIAN SURFACE IS TAKEN WITH REFERENCE TO CHARGE AND NOT CHARGE IS TAKEN WITH REFERENCE TO GAUSSIAN SURFACE.

perin

find the flux through the face of the cube, charge is q and area of each face is a.

suresan

Hi perin
To your question plaese see whether this idea helps or not.

If charge q is placed at one of the corners of the cube in free space then the net flux through the cube will be q/8e. Where e is the epsilon zero.
The flux through any of the three faces in which the charge is placed will be zero. And the flux through any of the other three faces which do not contain the charge will be q/24e.

nishantsingh89

for this type of questions, place the cubes identical to the one given
such that q is common vertex of all the cubes, and is at the centre of the bigger cube formed,
the bigger cube formed will have 8 identical small cubes, try imagining, place the cubes with q as common vertex, nd form a bigger cube)you'll see you need 8 cubes

now flux thru the big cube = q/e ( e---> epsilon)
now q is at centre of cube nd by symmetry, flux thru smaller cubes = q/8e
ans....

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