" If "oint_(s)[E.dS],[=0" over a surface,then "]

If oint_(s)vecE.vecdS=0 over a surface, then

Electric field is the electrostatic force per unit charge acting on a vanishingly small test charge placed at that point. It is a vector quantity and the electric field inside a charged conductor is zero. Electric flux phi is the total number of electric lines of force passing through a surface in a direction normal to the surface when the surface is placed inside the electric field. phi=ointvecE.vec(ds)=q/epsilon_0 If ointvecE.vec(ds)=0 over a surface, then

If oint_(S) vec(E ).vec(dS) = 0 over a surface then,

If underset s oint vecE.vec(ds)=0 over a surface, then

STATEMENT -1 : If ointvec(E).bar(ds) over a closed surface is negative,it means the surface encloses a net negative charge. STATEMENT-2 : We may have a Gaussian surface in which three lines enter and five field lines are coming out. STATEMENT-3 : The quantity ointvec(E).bar(ds) is independent of the charge distribution inside the surface.

STATEMENT -1 : If ointvec(E).bar(ds) over a closed surface is negative,it means the surface encloses a net negative charge. STATEMENT-2 : We may have a Gaussian surface in which three lines enter and five field lines are coming out. STATEMENT-3 : The quantity ointvec(E).bar(ds) is independent of the charge distribution inside the surface.

Let S be an imaginary closed surface enclosing mass m . Let dvecS be an element of area on S , the direction of dvecS being outward from S . Let vecE be the gravitational intensity at dvecS . We define phi=oint_(s)E*dS , the integration being carried out over the entire surface S .

Let S be an imaginary closed surface enclosing mass m . Let dvecS be an element of area on S , the direction of dvecS being outward from S . Let vecE be the gravitational intensity at dvecS . We define phi=oint_(s)E*dS , the integration being carried out over the entire surface S .