The space between two large parallel plates is filled with a material of uniform charge density $\rho$. Assume that one of the plates is kept at $x=0$. The potential at any point $x$ between these plates is given by (where $A$ and $B$ are constants):

The electric field intensity at a point between two parallel sheets with like charges of the same surface charge density $(\sigma)$ is:

$A$ conducting sphere of radius $R = 20 \ cm$ is given a charge $Q = 16 \ \mu C$. What is the electric field $\overrightarrow{E}$ at the centre?

$(a)$ Show that the normal component of the electrostatic field has a discontinuity from one side of a charged surface to another given by $(E_2 - E_1) \cdot \hat{n} = \frac{\sigma}{\varepsilon_0}$, where $\hat{n}$ is a unit vector normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\hat{n}$ is from side $1$ to side $2$.) Hence, show that just outside a conductor, the electric field is $\frac{\sigma \hat{n}}{\varepsilon_0}$. $(b)$ Show that the tangential component of the electrostatic field is continuous from one side of a charged surface to another.

The volume charge density of a sphere of radius $6 \, m$ is $2 \, \mu C \, m^{-3}$. The number of lines of force per unit surface area coming out from the surface of the sphere is $.... \times 10^{10} \, N C^{-1}$. [Given: Permittivity of vacuum $\epsilon_{0} = 8.85 \times 10^{-12} \, C^{2} N^{-1} m^{-2}$]

$A$ force of $10 \; N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of the capacitor is removed,then the force acting on that particle will be ...... $N$.