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ELECTRIC FLUX

Consider flow of a liquid with velocity v, through a small flat surface dS, in a direction normal to the surface. The rate of flow of liquid is given by the volume crossing the area per unit time v dS and represents the flux of liquid flowing across the plane. If the normal to the surface is not parallel to the direction of flow of liquid, i.e., to v, but makes an angle θ with it, the projected area in a plane perpendicular to v is δ dS cos θ. Therefore, the flux going out of the surface dS is v.ˆndS. For the case of the electric field, we define an analogous quantity and call it electric flux. We should, however, note that there is no flow of a physically observable quantity unlike the case of liquid flow.

Fig.1: Dependence of flux on the inclination θ between E and ˆn

In the picture of electric field lines described above, we saw that the number of field lines crossing a unit area, placed normal to the field at a point is a measure of the strength of electric field at that point. This means that if we place a small planar element of area ∆S normal to E at a point, the number of field lines crossing it is proportional (It will not be proper to say that the number of field lines is equal to E∆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.) to E ∆S. Now suppose we tilt the area element by angle θ. Clearly, the number of field lines crossing the area element will be smaller. The projection of the area element normal to E is ∆S cosθ. Thus, the number of field lines crossing ∆S is proportional to E ∆S cosθ. When θ = 90°, field lines will be parallel to ∆S and will not cross it at all (Fig. 1).

The orientation of area element and not merely its magnitude is important in many contexts. For example, in a stream, the amount of water flowing through a ring will naturally depend on how you hold the ring. If you hold it normal to the flow, maximum water will flow through it than if you hold it with some other orientation. This shows that an area element should be treated as a vector. It has a magnitude and also a direction. How to specify the direction of a planar area? Clearly, the normal to the plane specifies the orientation of the plane. Thus the direction of a planar area vector is along its normal.

How to associate a vector to the area of a curved surface? We imagine dividing the surface into a large number of very small area elements. Each small area element may be treated as planar and a vector associated with it, as explained before.

Fig.2

Notice one ambiguity here. The direction of an area element is along its normal. But a normal can point in two directions. Which direction do we choose as the direction of the vector associated with the area element? This problem is resolved by some convention appropriate to the given context. For the case of a closed surface, this convention is very simple. The vector associated with every area element of a closed surface is taken to be in the direction of the outward normal. This is the convention used in Fig. 2. Thus, the area element vector ∆S at a point on a closed surface equals ∆Sˆn where ∆S is the magnitude of the area element and ˆn is a unit vector in the direction of outward normal at that point.

We now come to the definition of electric flux. Electric flux ∆ϕ through an area element ∆S is defined by

ϕ = E.∆S = ES cosθ                         (1)

which, as seen before, is proportional to the number of field lines cutting the area element. The angle θ here is the angle between E and ∆S. For a closed surface, with the convention stated already, θ is the angle between E and the outward normal to the area element. Notice we could look at the expression E ∆S cosθ in two ways: E (∆S cos θ ) i.e., E times the projection of area normal to E, or E⊥∆S, i.e., component of E along the normal to the area element times the magnitude of the area element. The unit of electric flux is N.m2/C 

The basic definition of electric flux given by Eq. (1) can be used, in principle, to calculate the total flux through any given surface. All we have to do is to divide the surface into small area elements, calculate the flux at each element and add them up. Thus, the total flux φ through a surface S is

ϕ ~ Σ E.∆S                                            (2)

The approximation sign is put because the electric field E is taken to be constant over the small area element. This is mathematically exact only when you take the limit ∆S → 0 and the sum in Eq. (2) is written as an integral.

Example 1

A disk of radius 0.10 m is oriented with its normal unit vector at to a uniform electric field of magnitude (Fig. 22.7). (Since this isn’t a closed surface, it has no “inside” or “outside.” That’s why we have to specify the direction of in the figure.) (a) What is the electric flux through the disk? (b) What is the flux through the disk if it is turned so that is perpendicular to (c) What is the flux through the disk if is parallel to E S?

Fig.3: The electric flux Φ through a disk depends on the angle between its normal ˆn and the electric field E.

This problem is about a flat surface in a uniform electric field, so we can apply the ideas of this section. We calculate the electric flux using Eq. (22.1).

(a) The area is A=π(0.10 m)2=0.0134 m2 and the angle between E and A = Aˆn is ϕ=300, so from Eq. (22.1),

ΦE=EA cosϕ=(2.0×103 N/C)(0.0314 m2)(cos 300)=54 N.m2/C

(b) The normal to the disk is now perpendicular to E, ϕ=900, cos ϕ=0, and Φ=0

(c) The normal to the disk is parallel to E, so ϕ=0 and cos ϕ=1:

ΦE=EA cosϕ=(2.0×103 N/C)(0.0314 m2)(1)=63 N.m2/C

As a check on our results, note that our answer to part (b) is smaller than that to part (a), which is in turn smaller than that to part (c). Is all this as it should be?

Example 2

Fig.4
Assume the magnitude of the electric field on each face of the cube of edge L=1.11mL = 1.11 \, \text{m} in the figure below is uniform, and the directions of the fields on each face are as indicated. (Take E1=33.6N/CE_1 = 33.6 \, \text{N/C} and E2=25.5N/CE_2 = 25.5 \, \text{N/C}.) Find the net electric flux through the cube!

Solution:

To solve this, we need to find the net electric flux through the cube. the net flux can also be computed by summing the contributions of the electric flux through each face of the cube.

Φnet=E.A

Fig.5

Figure 5. shows the unit vectors ^n1 through ^n1 for each face; each unit vector points outward from the cube’s closed surface. 

The angle between E1 and ^n1 is 180°, then

ΦE1=E1.^n1A=E1L2 cos 1800=(33.6 N/C)(1.11 m)2=41.40 N.m2/C

The angle between E2 and ^n2 is 180°, then

ΦE2=E2.^n2A=E2L2 cos 1800=(25.5 N/C)(1.11 m)2=31.43 N.m2/C

The angle between E3 and ^n3 is 0°, then

ΦE3=E3.^n3A=E3L2 cos 00=(20.0 N/C)(1.11 m)2=24.64 N.m2/C

The angle between E4 and ^n4 is 0°, then

ΦE4=E4.^n4A=E4L2 cos 00=(15.0 N/C)(1.11 m)2=18.48 N.m2/C

The angle between E5 and ^n5 is 0°, then

ΦE5=E5.^n5A=E5L2 cos 00=(20.0 N/C)(1.11 m)2=24.64 N.m2/C

The angle between E6 and ^n6 is 0°, then

ΦE6=E6.^n6A=E6L2 cos 00=(20.0 N/C)(1.11 m)2=24.64 N.m2/C

The flux is negative on face 1, where E is directed into the cube, and positive on face 2, where E is directed out of the cube. The total flux through the cube is

ΦE=ΦE1+ΦE2+ΦE3+ΦE4+ΦE5+ΦE6

ΦE=[(41.40)+(31.43)+24.64+18.48+24.64+24.64] N.m2/C

ΦE=19.57 N.m2/C

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