Gauss's Law and its Applications Test - 1

CONCEPT

• Electric Flux: It is defined as the number of electric field lines passing through the perpendicular unit area.

• Electric Flux = (Φ) = EA_⊥ [E = electric field, A = perpendicular area]
• Electric flux (Φ) = EA cos θ [where θ is the angle between area plane and electric field]

• The flux is maximum when the angle is 0°
• Gauss Law: According to gauss’s law, total electric flux through a closed surface enclosing a charge is 1/ϵ₀ times the magnitude of charge enclosed.

\(i.e.~{{\mathbf{\Phi }}_{net}}=\frac{\left( {{Q}_{in}} \right)}{{{\epsilon }_{0}}}\)

\(i.e.~\oint \vec{E}\cdot d\vec{S}=\frac{{{Q}_{in}}}{{{\epsilon }_{0}}}\)

Where, Φ = electric flux, Q_in = charge enclosed the sphere, ϵ₀ = permittivity of space (8.85 × 10^-12 C²/Nm²), dS = surface area

EXPLANATION

According to gauss’s law,

The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ₀.

\(\oint \vec{E}\cdot d\vec{S}=\frac{{{Q}_{in}}}{{{\epsilon }_{0}}}\)

Electric flux on a closed surface only depends on the enclosed charge.

CONCEPT:

• Electric flux: The number of electric field lines passing through a surface area normally is called electric flux. It is denoted by Φ.

• The electric flux through a chosen surface is given by:

\({\rm{\Delta }}\phi = \vec E.{\rm{\Delta }}\vec S = E{\rm{\Delta }}Scos\theta \)

Where θ is the angle between the electrical field and the positive normal to the surface.

EXPLANATION:

• The number of electric lines of force passing through unit area is called electric flux. So option 1 is correct.
• The electric charge per unit area is called electric surface charge density of that surface.
• The space or region around the electric charge in which electrostatic force can be experienced by other charged particles is called an electric field by that electric charge.

EXTRA POINTS:

• Gauss’s Law: It states that the net electric field through a closed surface equals the net charge enclosed by the surface divided by ϵ0.

\(\oint \vec E.\overrightarrow {ds} = \frac{{{q_{inside}}}}{{{ϵ_0}}}\)

Where E = electric field, ds = small area, qinside = the total charge inside the surface, and ϵ0 = the permittivity of free space.

CONCEPT:

Electric field intensity: 

• It is defined as the force experienced by a unit positive test charge in the electric field at any point.

\(⇒ E=\frac{F}{q_{o}}\)    

Where E = electric field intensity, qo = charge on the particle

Electric field intensity due to an infinitely long straight uniformly charged wire:

• The direction of the electric field at every point must be radial (outward if λ > 0, inward if λ < 0).

\(⇒ E=\frac{λ}{2\piϵ_or}\)

Where λ = linear charge density, ϵo = permittivity, and r = distance of the point from the wire

EXPLANATION:

We know that the electric field intensity due to an infinitely long straight uniformly charged wire is given as,

\(⇒ E=\frac{λ}{2\piϵ_or}\)

⇒ E ∝ λ      ----(1)

Where λ = linear charge density, ϵo = permittivity, and r = distance of the point from the wire

• Hence, option 2 is correct.

Concept:

Gauss's law, also known as Gauss's flux theorem, is a law relating the distribution of electric charge to the resulting electric field.

• The surface under consideration may be a closed one enclosing a volume such as a spherical surface.
• Flux through the surface is the number of field lines flowing through that surface.
• Field lines are imaginary lines that depict the strength and direction of the field.
• Since the electric field will be generated by electric charge itself, and hence its direction should depend on the electric charge that is present inside it.

Spherical shell:

• All the charge resides on the surface and not inside it.
• There is not even a quantum of charge inside the shell. So, when we construct any closed surface inside the shell, that surface encloses no charge.
• Since inside the shell, it doesn’t enclose any charge then the electric field inside the shell is zero always.

Concept:

By Gauss Law:

\({\phi _E} = \oint \vec E \cdot \overrightarrow {dA} = \frac{{{Q_{enc}}}}{{{\epsilon_0}}}\)

Analysis:

Electric Field due to spherical shell shows the following behavior:

Electric flux:

\({{\int\ \!\!\!\!\!\int}\ \mkern-21mu \bigcirc} {E.ds = \frac{Q}{\epsilon }}\)

E has a unit of V/m and ds has 'a' unit of m2. So the S.I. unit will be:

Since the net flux is the total field passing through an area. We define the unit as:

\(\frac{V}{m}×{m^2}=V-m\)

Electrical flux has SI units of voltmeters (V-m), or, equivalently, Newton-meters squared per coulomb (N m2 C-1).

Important Points

CONCEPT:

Electric flux:

• It is defined as the number of electric field lines associated with an area element.
• Electric flux is a scalar quantity.
• The SI unit of the electric flux is N-m2/C.
• If the electric field is E and the area is A then the electric flux associated with the area is given as,
  \(⇒ ϕ = \vec{E}.\vec{A}\)
  ⇒ ϕ = EA cos θ
  Where θ = angle between the surface and the electric field

EXPLANATION:

• If the electric field is E and the area is A then the electric flux associated with the area is given as,
  ⇒ ϕ = EAcosθ     ---(1)
  Where θ = angle between the surface and the electric field
• By equation 1 it is clear that the electric flux depends on the electric field intensity, area, and the angle between the surface and the electric field. Hence, option 4 is correct.

CONCEPT:

Electric flux:

• It is defined as the number of electric field lines associated with an area element.
• Electric flux is a scalar quantity.
• The SI unit of the electric flux is N-m2/C.
• If the electric field is E and the area is A then the electric flux associated with the area is given as,

\(⇒ ϕ = \vec{E}.\vec{A}\)

⇒ ϕ = EAcosθ

Where θ = angle between the surface and the electric field

EXPLANATION:

If the electric field is E and the area is A then the electric flux associated with the area is given as,

⇒ ϕ = EAcosθ     -----(1)

Where θ = angle between the surface and the electric field

• Here the plane of the ring is parallel to the electric field so the angle between the surface and the electric field will be 90°.

So by equation 1, the electric flux associated with the ring will be,

⇒ ϕ = EA×cos90

⇒ ϕ = 0

• Hence, option 1 is correct.

Concept:

Gauss’s Law: Total electric flux through a closed surface is 1/εo times the charge enclosed in the surface i.e.

\({\rm{\Phi }} = \frac{q}{{{\epsilon_o}}}\)

But we know that Electrical flux through a closed surface is:

\(\oint \vec E \cdot \overrightarrow {ds} \)

\(\therefore\oint \vec E \cdot \overrightarrow {ds} = \frac{q}{{{\epsilon_o}}}\)

E = electric field

q = charge enclosed in the surface and

εo = permittivity of free space.

The electric field at a point due to an infinite sheet of charge is:

\(E=\frac{\sigma }{2{{\epsilon }_{0}}}\)

Where ϵ_o = Absolute electrical permittivity of free space, E = Electric field, and σ = surface charge density. 

Thus, the field is uniform and does not depend on the distance from the plane sheet of charge. Therefore option 3 is correct.

The correct answer is option 1) i.e. \(\frac{5q}{\epsilon_0}\)

CONCEPT:

• Gauss's Law for electric field: It states that the total electric flux emerging out of a closed surface is directly proportional to the charge enclosed by this closed surface.

It is expressed as:

\(ϕ = \frac{Q}{ϵ_0}\)

Where ϕ is the electric flux, Q is the charge enclosed in the closed surface and ϵ0 is the electric constant.

EXPLANATION:

• According to Gauss law, the flux depends on the total charge enclosed by the surface.

Here, the total charge (Q) = 2q + 3q = 5q

Therefore, electric flux \(ϕ = \frac{5q}{ϵ_0}\)