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
$\frac{1}{2}{\varepsilon _0}{E^2}Ad$
${\varepsilon _0}{E}Ad$
$\frac{1}{2}{\varepsilon _0}{E^2}$
${E^2}Ad/{\varepsilon _0}$
If the electric potential of the inner shell is $10\,V$ and that of the outer shell is $5\,V$ , then the potential at the centre will be.....$V$
In the given figure, three capacitors $C_1, C_2$ and $C_3$ are joined to a battery, with symbols having their usual meanings, the correct conditions will be
The work done required to put the four charges together at the corners of a square of side $a$ , as shown in the figure is
The potential $V$ is varying with $x$ and $y$ as $V = \frac{1}{2}({y^2} - 4x)\,volts$ The field at $(1\,m,\,1\,m)$ 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