Question

Consider a charge placed outside an empty(not containing any charge inside it)gaussian surface then as the surface doesnt enclose any charge the field inside gaussian surface should be zero because integral E.da=Qenclosed=0.but there is field present around the charge.so can anyone please explain me this contradiction.im confused about it

Answer 1

i once had this same qn in my mind...in a book i saw this...now say to calculate electric flux the surface integral E.dS...whose electric field will u consider?

Answer 2

well i wanted to know that accordinding to gauss law when 1 and two are removed and only charge 3 is present then field inside the sphere should be zero but there is always e field present in the vicinity of electric charge.plzz explain in detail how is this possible?

Answer 3

no no...first the ans to my qn was all the THREE...remember gausss law the value of q in the gauss law eqn is ONLY the charges inside the sphere..but E in the eqn is due to ALL the three....and reg ur qn wat i think is...gauss law is a mathemetical trick...it is used mainly to find electric field due to diff shapd objects...we construct gaussian sphere in a way to get the req ans...and i think its not as fundamental as coulombs law...but rather a technique

Answer 4

@ujjwal do u have any better answers than me?

Answer 5

@imagreencat @Vaidehi09

Answer 6

was the ans that ALL THREE CHARGES given in the book or is it your own answer???
as far as i know i guess gauss law is very fundamental like couloumbs and using gauss law only prof walter in his lecture proved that field insidie charged spherical shell is zero though the charge is present outside it in spherical manner rather that just adding the effect of field due to all charges present outside sphere.i hope u got my point of why i guess your approach is not so good according to me

Answer 7

it was given in the book...even in the spherical shell problem if we consider the electricc field due to all charges the ans will be zero because the charges inside the inner sphere is zero...what made me to think that gaussslaw isnt fundamental(dont take it literally) because GAUSSLAW WONT HOLD IF COULOMBS LAW DOSNT HOLD...i maybe wrong...if iam ill corect it:) pls wait for @ujjwal hes good in electricity

Answer 8

ohh.well but i do know and agree that summing the effect of all charges would make field inside it zero but walter dint use that approach rather he used gauss law to prove it.if your approach was correct@mathavraj then i guess walter shouldnt have used gauss law rather used superposition principle

Answer 9

no i dint use superposition either..i just used the formula of gausslaw only....in which q is the charge inside the gaussian surface

Answer 10

@sami-21 @Muskan

Answer 11

@mathslover

Answer 12

@theyatin @TheViper

Answer 13

huhuhu ha ha ha tell my lord why you called ginny here???

Answer 14

Are you talking about electric flux? The electric flux through a closed surface is zero as long as the charge is outside the closed surface.\[\phi=\frac{q}{\epsilon_0}\]Where q is charge enclosed by the surface.
This one explains much better. http://facultyfiles.deanza.edu/gems/lunaeduardo/ElectricFluxandGausssLawinEM.pdf

Answer 15

why r u mixing up things when we say Gauss law it is just about the flux, it has no relation with what force u r experience, Flux is zero bcoz solid angle is zero and not E field is zero
they both are in dot product any thing zero will blew it up!!!!!
yhe to chillar doubt tha

Answer 16

it is the property of charge to come out on the surface of hollow conductor so basically here charge remains on the surface of gaussian surface and then we calculate electric field/flux, still charge has its electric field so nothing is contradictory ....

Answer 17

i just used conductor as an example to demonstrate the hollow sphere.. i mean it is the property of charge to get uniformly distributed over a surface if it is placed inside a hollow conductor..that's it ..same is the case while considering gaussian surface and deriving field formula

Answer 18

It is simple.Since there is no charge inside the Gaussian surface the total number of field lines entering the surface must exit. So there is flux inside the surface but the "NET" flux is zero.