Test: Electric Flux & Gauss Law

► Electric field (E) = 26i - k
► A = 20k
Electric flux = electric field.area
= E.A
= (26i - k)*(20k)
= (26*0 -1)*20
= -20 units

Net electric flux (Φ_Net) through the cubic surface is given by,

Where, ∈₀ = Permittivity of free space

= 8.854 × 10⁻¹² N⁻¹C² m⁻²

q = Net charge contained inside the cube = 2.0 μC = 2 × 10⁻⁶ C

∴

= 2.2 × 10⁵ N m² C⁻¹

The net electric flux through the surface is 2.2 ×10⁵ N m²C⁻¹.

For constant electric field

Electric flux is given by  since amount of charge not depends on size and shape so by making radius <  double the amount of charge remain same so electric flux remain same.

By definition of Solid Angle,

Where, Ω is represented in Steradians.

Electric flux is a scalar quantity. Electric Flux is dot/scalar product of magnetic field vector and area vector. And the result of a dot/scalar quantity is always a scalar quantity. So electric flux is also a scalar quantity 

► Area = 20i since it is in the Y-Z plane
► E = 6i + 3j + 4k
► E.S = Flux
= 20*6 = 120 units

Line charge density = λ
Radius, r = R
Now,  λdl=dθ
For total charge enclosed by radius R, ₀∫^R Q = ₀∫^R λdl
Q_inc = λR
Now, ϕ = Q_inc/ε_o    {from Gauss law}
= λR/ε₀

Flux = E. A
⇒ F = Eq
⇒ E= F/q= N/C
⇒ Flux= N/C × m²
= Nm²C^-1

Electric flux density is the charge per unit area. Hence it is a function of charge and not any of the other values.

The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. In other words, electric flux per unit volume leaving a point (vanishing small volume), is equal to the volume charge density

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By definition,

For maximum, E.dA must be maximum. This is possible when E || dA.

A point charge is single dimensional. The three dimensional imaginary enclosed surface of a point charge will be sphere.

By Gauss Law,

Electric Field due to spherical shell shows the following behavior:

We can assume a similar vessel, such that the charge lies at the interfaces of both the vessels. Now, by Gauss law, all the electric flux must pass through these two vessels. As they are kept symmetrically with respect to the charge equal amount of flux would pass through them

From both vessels,

For each vessel,

We know that the flux linked with a Gaussian surface is given as:

Φ = E.ds = q/ε₀

Here,

q = 8.85 x 10^-8 C

ε₀ = 8.85 x 10^-12 F/m

So,

Φ = 8.85x10^-8/ 8.85x10^-12

Thus, flux will be

Φ = 10^-4 Nm²/C

Hence,

If the radius of the spherical shell is double then also flux associated with spherical surface remain unchanged because the charge enclosed in system always remains unchanged.

A line charge can be visualized as a rod of electric charges. The three dimensional imaginary enclosed surface of a rod can be a cylinder.

We calculate this using spherical symmtery in Gauss Law

For spherical Shell,