Doubt from BPHCT 133 , block 3 , page no 22

[Chanda Kumari]

Respected teacher's please kindly explain me equation
10.21 to 10.25 . On page no 22 of block3 of BPHCT 133

[SR Jha]

Dear Chanda Kumari,

It seems you are facing difficulty in understanding the steps of derivation from Eq. (10.21) to Eq. (10.25) in the Section 10.4 on ‘Physical interpretation of Bound Charge density’ in Unit 10, Block 3 of the Course. As has been mentioned in the course material (Se. 10.4, Unit 10), Eq. (10.21) is obtained by integrating Eq. (10.19). The right hand side of Eq. (10.21) is obtained if we write  because  is a unit vector perpendicular to the surface area A.

Since Eq. (10.21) gives the net outflow of the charge from the given volume due to polarisation, the volume under consideration will be left with equal and opposite charge; that is, –Q. that is how we get Eq. (10.22a).

The first part of Eq. (10.22b) comes from the definition of charge contained in a volume (dV) in terms of volume charge density and the volume element dV under consideration. The second part of Eq. (10.22b) is simply substituting for –Q from Eq. (10.22a) in the first part of Eq. (10.22b).

Eq. (10.23) is the well known Gauss’s divergence theorem relating surface and volume integrals; you should refer to Unit 4 of this course to understand the physical significance of this theorem.

Lastly, Eqs. (10.24) and (10.25) are simple algebra of substituting earlier Eqs. in Eq. (10.23) to obtain the desired result.

Hope it clarifies your difficulty in understanding.