Two semi-infinite conducting sheets are placed at right angles to each other as shown in the figure. A point charge of +𝑄 is placed at a distance of 𝑑 from both sheets. The net force on the charge is $$\frac{Q^2}{4{\mathrm{πε}}_0}\frac{\overrightarrow K}{d^2}$$ , where $$\overrightarrow K$$ is given by

A perfectly conducting metal plate is placed in x-y plane in a right handed coordinate system. A charge of $$+32{\mathrm{πε}}_0\sqrt2$$ columbs is placed at coordinate (0, 0, 2). $${\mathrm\varepsilon}_0$$ is the permittivity of free space. Assume $$\widehat i,\;\widehat j,\;\widehat k$$ to be unit vectors along x, y and z axes respectively. At the coordinate $$\left(\sqrt2,\sqrt2,0\right)$$, the electric field vector $$\overrightarrow E$$ (Newtons/Columb) will be

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