The charge $q$ on a capacitor varies with voltage as shown in figure. The area of the triangle $AOB $ represents
electric field between the plates
electric flux between the plates
energy density
energy stored by the capacitor
$Assertion\,(A):$ A charge $q$ is placed on a height $h / 4$ above the centre of a square of side b. The flux associated with the square is independent of side length.
$Reason\,(R):$ Gauss's law is independent of size of the Gaussian surface.
An electric field $\overrightarrow{\mathrm{E}}=4 \mathrm{x} \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}\; \mathrm{N} / \mathrm{C}$ passes through the box shown in figure. The flux of the electric field through surfaces $A B C D$ and $BCGF$ are marked as $\phi_{I}$ and $\phi_{\mathrm{II}}$ respectively. The difference between $\left(\phi_{\mathrm{I}}-\phi_{\mathrm{II}}\right)$ is (in $\left.\mathrm{Nm}^{2} / \mathrm{C}\right)$
A point charge causes an electric flux of $-1.0 \times 10^{3}\; N\;m ^{2} / C$ to pass through a spherical Gaussian surface of $10.0\; cm$ radius centred on the charge.
$(a)$ If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
$(b)$ What is the value of the point charge?
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^{3} \;Nm ^{2} / C .$
$(a)$ What is the net charge inside the box?
$(b)$ If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?
As shown in figure, a cuboid lies in a region with electric field $E=2 x^2 \hat{i}-4 y \hat{j}+6 \hat{k} \quad N / C$. The magnitude of charge within the cuboid is $n \varepsilon_0 C$. The value of $n$ is $............$ (if dimension of cuboid is $1 \times 2 \times 3 \;m ^3$ )