A dielectric slab with 500 mm x 500 mm cross-section is 0.4 m long. The slab is subjected to a uniform electric field of $$E=6{\widehat a}_x+8{\widehat a}_y$$ kV /mm. The relative permittivity of the dielectric material is equal to 2. The value of constant $$\varepsilon_0$$ is $$8.85\times10^{-12}\;F/m$$ . The energy stored in the dielectric in Joules is

A capacitor is made with a polymeric dielectric having an $$\varepsilon_0$$ of 2.26 and a dielectric breakdown strength of 50 kV/cm. The permittivity of free space is 8.85 pF/m. If the rectangular plates of the capacitor have a width of 20 cm and a length of 40 cm, then the maximum electric charge in the capacitor is

$$$F\left(x,y\right)=\left(x^2\;+\;xy\right)\;{\widehat a}_x\;+\;\left(y^2\;+\;xy\right)\;{\widehat a}_y$$$. Its line integral over the straight line from (x, y)=(0,2) to (2,0) evaluates to

A capacitor consists of two metal plates each $$500 \times 500\,\,m{m^2}$$ and spaced $$6$$ $$mm$$ apart. The space between the metal plates is filled with a glass plate of $$4$$ $$mm$$ thickness and a layer of paper of $$2$$ $$mm$$ thickness. The relative permittivity of the glass and paper are $$8$$ and $$2$$ respectively. Neglecting the fringing effect, the capacitance will be (Given that $${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,F/m$$ )