A parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and area of each plate is $A$ , the energy stored in the capacitor is
$\frac{1}{2}{\varepsilon _0}{E^2}Ad$
${\varepsilon _0}{E}Ad$
$\frac{1}{2}{\varepsilon _0}{E^2}$
${E^2}Ad/{\varepsilon _0}$
Two condensers $C_1$ and $C_2$ in a circuit are joined as shown in figure. The potential of point $A$ is $V_1$ and that of $B$ is $V_2$. The potential of point $D$ will be
In the figure shown, the potential difference between points $A$ and $B$ is......$V$
A parallel plate capacitor is charged to a potential difference of $100\,V$ and disconnected from the source of $emf$. A slab of dielectric is then inserted between the plates. Which of the following three quantities change?
$(i)$ The potential difference $(ii)$ The capacitance $(iii)$ The charge on the plates
The electric potential $V$ at any point $(x,y,z)$ in space is given by equation $V = 4x^2\,volt$ where $x, y$ and $z$ are all in metre. The electric field at the point $(1\,m, 0, 2\,m)$ in $V/m$ is
If $\vec E = \frac{{{E_0}x}}{a}\hat i\,\left( {x - mt} \right)$ then flux through the shaded area of a cube is