Solution:

Ans. Try yourself, Similar to Q. 3 Short Answer Type Questions-II

3

Long Answer Type Questions

[AI Q. 1. (i) Define electric flux. Is it a scalar or a vector quantity?

A point charge $q$ is at a distance of $d / 2$ directly above the centre of a square of side $d$, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.

(ii) If the point charge is now moved to a distance ’ $d$ ’ from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected. R&A[CBSE Delhi OD 2018]

Ans. (i) Definition of electric flux Stating scalar/ vector

Gauss’s law

Derivation of the expression for electric flux

(ii) Explanation of change in electric flux the dot product of electric field and area vector ovet that surface.

Alternatively $\phi=\oint_{s} \vec{E} \cdot \overrightarrow{d S}$

Also accept

Electric flux, through a surface equals the surface integral of the electric field over that surface. $1 / 2$ It is a scalar quantity.

Constructing a cube of side ’ $d$ ’ so that charge ’ $q$ ’ gets placed within of this cube (Gaussian surface )

According to Gauss’s law the Electric flux

$$ \begin{aligned} \phi & =\frac{\text { Charge enclosed }}{\varepsilon_{0}} \ & =\frac{q}{\varepsilon_{0}} \end{aligned} $$

This is the total flux through all the six faces of the cube. Hence electric flux through the square

$$ \frac{1}{6} \times \frac{q}{\varepsilon_{0}}=\frac{q}{6 \varepsilon_{0}} $$

(ii) If the charge is moved to a distance $d$ and the side of the square is doubled the cube will be constructed to have a side $2 d$ but the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore the flux through the square will remain the same as before. [Deduct 1 mark if the student just writes No change /not affected without giving any explanation.] 1+1

[CBSE Marking Scheme 2018]