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# Electric Flux Assignment Help

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Electrostatics - Electric Flux

**Electric Flux**

Electric flux over an area in an electric field represents the total number of field lines crossing this area.

We know that the number of field lines crossing a unit area placed normal to the field at a point is a measure of strength of electric field E at that point. If we place a small planar element of area** Δ S **normal to E at this point, number of electric filed lines crossing this area element is proportional to **E ( Δ S)** note that it is not proper to say that number of field lines crossing the area is equal to **E ( Δ S)**. The number of field lines is after all a matter of how many field lines we choose to draw. What is physically significant is the relative number of field lines crossing a given area at different points.

If we tilt the area element by angle o [or we tilt E w. area element by angle θ fig.1 (c) the number of filed lines crossing the area will be smaller. As projection of area element normal to E is ** Δ S** cos θ (or component of E normal of electric filed lines crossing area ?S is proportional to **E Δ S** cos θ. This is shown in fig 1 (c)

Hence electric flux ** Δ ∅ **through an area element ** Δ S** in an electric field E is defined as

**Δ ∅ = E. Δ S = E ( Δ S) cos θ**

This is proportional to the number of field lines cutting the area element.

Here θ is smaller angle between E and ** Δ S**. for a closed surface, θ is the angle between E and outward normal to the area element.

**Now, Δ ∅ = E . Δ S = E Δ S cos θ = E ( Δ S cosθ),**

E times the projection of area normal to E.

**Also, Δ ∅ = E . Δ S = E Δ S cos θ = (E cos θ) Δ S,**

Component of E along normal to the area element times the magnitude of area element when E is normal to area element **θ = 0**. electric flux is maximum when E is along the area element, **θ = 90** electric flux is zero. When** θ > 90 cos θ** is negative therefore electric flux is negative.

To calculate the total electric flux through any given surface, we have to divide the given surface into small area elements calculate the flux at each element and add them up. Therefore, total electric flux through a surface S is

**∅ε ≈ ∑ E. Δ S**

However when we take the limit ** Δ S --> O**, the summation can be written as integral and we obtain the exact value of electric flux

**∅ε = ƒE. Ds**

The circle on the integral sing indicates that the surface of integration is a closed surface.

Electric flux is a scalar quantity.

Units of **∅ε** = unite of E x unit of **S = NC**^{ -1 }x m^{2} = Nm^{2} C^{-1}

Dimensional formula of **∅ε**;

**[MLT**^{-2}] (L^{2}) [AT]^{ -1} = [M_{1} L^{3 }T^{-3} A^{-1}]

Electric Flux Assignment Help, Electric Flux Homework Help, Electric Flux Tutors, Electric Flux Solutions, Electric Flux Tutors, Electrostatics Help, Physics Tutors, Electric Flux Questions Answers

**Electric Flux**

We know that the number of field lines crossing a unit area placed normal to the field at a point is a measure of strength of electric field E at that point. If we place a small planar element of area

**Δ S**normal to E at this point, number of electric filed lines crossing this area element is proportional to

**E ( Δ S)**note that it is not proper to say that number of field lines crossing the area is equal to

**E ( Δ S)**. The number of field lines is after all a matter of how many field lines we choose to draw. What is physically significant is the relative number of field lines crossing a given area at different points.

If we tilt the area element by angle o [or we tilt E w. area element by angle θ fig.1 (c) the number of filed lines crossing the area will be smaller. As projection of area element normal to E is

**Δ S**cos θ (or component of E normal of electric filed lines crossing area ?S is proportional to

**E Δ S**cos θ. This is shown in fig 1 (c)

Hence electric flux

**Δ ∅**through an area element

**Δ S**in an electric field E is defined as

**Δ ∅ = E. Δ S = E ( Δ S) cos θ**

This is proportional to the number of field lines cutting the area element.

Here θ is smaller angle between E and

**Δ S**. for a closed surface, θ is the angle between E and outward normal to the area element.

**Now, Δ ∅ = E . Δ S = E Δ S cos θ = E ( Δ S cosθ),**

E times the projection of area normal to E.

**Also, Δ ∅ = E . Δ S = E Δ S cos θ = (E cos θ) Δ S,**

Component of E along normal to the area element times the magnitude of area element when E is normal to area element

**θ = 0**. electric flux is maximum when E is along the area element,

**θ = 90**electric flux is zero. When

**θ > 90 cos θ**is negative therefore electric flux is negative.

To calculate the total electric flux through any given surface, we have to divide the given surface into small area elements calculate the flux at each element and add them up. Therefore, total electric flux through a surface S is

**∅ε ≈ ∑ E. Δ S**

However when we take the limit

**Δ S --> O**, the summation can be written as integral and we obtain the exact value of electric flux

**∅ε = ƒE. Ds**

The circle on the integral sing indicates that the surface of integration is a closed surface.

Electric flux is a scalar quantity.

Units of

**∅ε**= unite of E x unit of

**S = NC**

^{ -1 }x m^{2}= Nm^{2}C^{-1}

Dimensional formula of

**∅ε**;

**[MLT**

^{-2}] (L^{2}) [AT]^{ -1}= [M_{1}L^{3 }T^{-3}A^{-1}]