The spatial distribution of the electric field due to charges $(A, B)$ is shown in figure. Which one of the following statements is correct
$A$ is $+ve$ and $B\, -ve$ and $|A| > |B|$
$A$ is $-ve$ and $B\, +ve; |A| = |B|$
Both are $+ve$ but $A > B$
Both are $-ve$ but $A > B$
The electric field in a region is given by $\overrightarrow{ E }=\frac{2}{5} E _{0} \hat{ i }+\frac{3}{5} E _{0} \hat{ j }$ with $E _{0}=4.0 \times 10^{3}\, \frac{ N }{ C } .$ The flux of this field through a rectangular surface area $0.4 \,m ^{2}$ parallel to the $Y - Z$ plane is ....... $Nm ^{2} C ^{-1}$
A square surface of side $L$ metres is in the plane of the paper. A uniform electric field $\vec E(V/m) $, also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is
The electric flux passing through the cube for the given arrangement of charges placed at the corners of the cube (as shown in the figure) is
An electric line of force in the $xy$ plane is given by equation ${x^2} + {y^2} = 1$. A particle with unit positive charge, initially at rest at the point $x = 1,\;y = 0$ in the $xy$ plane
A point charge $+10\; \mu \,C$ is a distance $5 cm$ directly above the centre of a square of side $10 \;cm ,$ as shown in Figure. What is the magnitude of the electric flux through the square?