From the study of subatomic particles to the laws of motion, Physics Topics offer insights into the workings of the world around us.
How do you find the direction of a Area Vector?
Area vector: In different cases of physics, surface area is treated as a vector. The magnitude of the area is represented by the length of the vector and the direction of the area vector is given by the outward drawn normal to the closed surface. Sup-pose, in Fig., dS is a small surface area on the surface S. A normal is drawn in the outward direction on the surface at dS. So d\(\vec{S}\) is an area vector.
Electric flux: Electric flux linked with a surface gives an idea of the number of field lines passing through the surface.
The number of electric field lines passing normally through a surface is called the electric flux through the surface. It is denoted by the symbol ϕ. It is a scalar quantity.
Let us consider a small area element d\(\vec{S}\) on a surface S in an electric field of inten-sity \(\vec{E}\) [Fig.]. Let θ be the angle between \(\vec{E}\) and d\(\vec{S}\).
Therefore, electric flux passing through the surface d\(\vec{S}\),
dϕ = \(\vec{S}\) ᐧ d\(\vec{S}\) = EdS cosθ
= (E cosθ)dS = EndS ……… (1)
Here, En = Ecosθ = normal component of electric field inten-sity.
As the whole surface is the sum of a large number of such small area elements, the electric flux passing through the surface S is given by
where represents the integral taken over the whole surface.
Special cases:
If an electric field E is normal to a surface, then θ = 0°. So the electric flux linked with the surface,
ϕ = \(\int_S\) Ed cos 0° = ES (maximum)
ii) If an electric field \(\vec{E}\) is parallel to a surface, then θ = 90°. So the electric flux linked with the surface,
ϕ = \(\int_S\) EdS cos90° = 0
Closed surface: Now, let us consider a definite volume enclosed by a closed surface S [Fig.]. Naturally, the electric flux linked with this surface S will be,
Here, the symbol \(\oint_S\) stands for integration across all the surface elements d\(\vec{S}\) which has different directions around the entire closed surface S. Moreover, \(\vec{E}\) may be different at different points on the surface. So, none of θ and E are constant when the entire surface is taken into account. As a result, the integral in equation (3) is often very hard to compute. However, specific symmetries in connection with some special systems help us to tackle the integral effectively.
Positive electric flux: The electric flux linked with a closed surface is taken as positive when the electric field vector is outwardly directed [Fig.(a)]. Here, the volume acts as a source of field lines.
Negative electric flux: The electric flux linked with a closed surface is taken as negative when the electric field vector is inwardly directed [Fig. (b)]. Here, the volume acts as a sink of field lines.
Unit and dimension of electric flux: In SI, unit of electric flux is N ᐧ m2 ᐧ C-1 or V ᐧ m.
Dimension of electric flux
= dimension of electric intensity × dimension of area
= MLT-3I-1 × L2 = ML3T-3I-1
Solid Angle: If a surface is placed at any distance in front of a point, then it subtends a solid angle at that point. In Fig., this surface of area S subtends a solid angle Ω at the point O.
Unit solid angle: The solid angle subtended by an area r2 on a sphere of radius r, at the centre of the sphere [Fig. (b)], is called a unit solid angle. This unit of solid angle is called stera- dian (sr).
The solid angle subtended by a surface of area S on a sphere of radius r, at the centre of the sphere [Fig.], is given by,
So, solid angle is a dimensionless quantity. Steradian is a dimensionless unit.
The solid angle at the centre of a sphere of radius r due to the entire surface of the sphere \(=\frac{\text { area of the entire surface }}{r^2}\) = \(\frac{4 \pi r^2}{r^2}\) = 4π
If the surface is a closed surface (not necessarily a sphere), then also it subtends a solid angle of 4π at any of its internal points.
Now, let us take a surface dS at a distance r from point O [Fig. (d)], In general, dS is not normal to the radius vector \(\vec{r}\) .If θ be the angle between d\(\vec{S}\) and \(\vec{r}\), then the component of dS along the normal to r = dS cosθ.
So, the solid angle subtended at O by the surface dS is,
dω = \(\frac{d S \cos \theta}{r^2}\)
Obviously, if a closed surface of any shape surrounds the point O, then the solid angle subtended at O by that closed surface,
\(\oint d \omega\) = \(\oint \frac{d S \cos \theta}{r^2}\) = 4π