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

  • A

    $\frac{1}{2}{\varepsilon _0}{E^2}Ad$

  • B

    ${\varepsilon _0}{E}Ad$

  • C

    $\frac{1}{2}{\varepsilon _0}{E^2}$

  • D

    ${E^2}Ad/{\varepsilon _0}$

Similar Questions

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