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No, Gauss’s law will be applicable for those fields for which they follow inverse square law.

Energy is stored in the volume of electric field.

As electric field inside charged sphere is zero (E = 0), therefore emergent flux from Gaussian surface will be zero.
or ϕ = 0 but, ϕ= \(\frac{q}{\varepsilon_0}\) ⇒ q = ϕε₀ = 0
It means, charge inside surface will be zero. Thus, the whole charge Q will be distributed on its surface.

There is equilibrium inside positive and negative charges of soap bubble. When it is given negative charge, then the position of equilibrium start to change. To achieve this equilibrium, charges try to repel each other. Thus the surface area of bubble increases. Due to increase in surface area, the bubble increases in size.

Volume charge density: When the charge density is throughout the volume of a body, then the charge per unit volume is called volume charge density and it is represented by ρ.
∴ ρ = \(\frac{q}{V}\)
Hence the unit of ρ is Cm^-3
Example: If a charge q is uniformly distributed in the whole volume of a sphere of radius R, then \(\rho =\frac { q }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } =\frac { 3q }{ 4\pi { R }^{ 3 } } \)

Surface charge density: When the charge is distributed on a plane or curved surface, then charge per unit area is called surface charge density of charge and represented by σ.
∴ σ = \(\frac{q}{A}\)
Where q is charge on surface of area A
∴ Unit of σ = Cm^-2
Example: If q charge is distributed uniformly on sphere of radius R, then surface charge density of the sphere \(\sigma=\frac{q}{4 \pi R^{2}}\).

In both conditions, the ratio of electric flux will be equal.

Linear charge density: When the charge density is along a line, then the quantity of charge per unit length is called linear charge density and it is represented by λ.
λ = \(\frac{q}{l}\), where q is charge uniformly distributed on length l.
Unit of λ = Cm^-1
Example: If q charge is uniformly distributed on a ring of radius R, then the linear charge density of ring will be \(\frac{q}{2 \pi R}\)

Change in electric field
= E₁ – E₂
\(=\frac{\sigma}{2 \varepsilon_{0}}-\left(-\frac{\sigma}{2 \varepsilon_{0}}\right)=\frac{\sigma}{\varepsilon_{0}}\)

Uniform charged insulating sphere has electric field intensity zero at centre and infinity.

If the area of elemental surface \(\overrightarrow{d s}\) is at 90° to electric field, then emergent out flux will be zero.

(c) Zero

Due to dipole, there is no area from which flux is passing. Thus the leaving out flux = 0.

(c) zero
Since electric field is acting along, plane of the square, so the angle between area vector \(\vec A\) and \(\vec E\) is 90°. These give ϕ = EAcosθ = EAcos90 = 0.

We will close the window of the car. So that entire charge will lie on the surface of car.