AIPMT | 2015Gauss LawEasy
Question
The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a shepere of radius ′a′ centred at the origin of the field, will be given by :
Options
A.A ε0 a2
B.4 πε0 Aa3
C.ε0 Aa3
D.4 πε0 Aa2
Solution
Flux linked with sphere = 
since electric field is radial. It is always perpendicular
to the surface.
so φ - Ar. (4πr2)
φ - A (a)(4πr2) (as r = a)
φ - A4πa3)
Now according to gauss law
φ =
⇒ qin = φ.∈0
so qin = A4πa3∈0
Hence option (2)
since electric field is radial. It is always perpendicular
to the surface.
so φ - Ar. (4πr2)
φ - A (a)(4πr2) (as r = a)
φ - A4πa3)
Now according to gauss law
φ =
⇒ qin = φ.∈0so qin = A4πa3∈0
Hence option (2)
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