The electric field due to a uniformly charged sphere of radius $R$ as a function of the distance $r$ from its centre is represented graphically by
A thin infinite sheet charge and an infinite line charge of respective charge densities $+\sigma$ and $+\lambda$ are placed parallel at $5\,m$ distance from each other. Points $P$ and $Q$, are at $\frac{3}{\pi} m$ and $\frac{4}{\pi} m$ perpendicular distance from line charge towards sheet charge, respectively. $E_P$ and $E_Q$ are the magnitudes of resultant electric field intensities at point $P$ and $Q$, respectively. If $\frac{E_p}{E_Q}=\frac{4}{a}$ for $2|\sigma|=|\lambda|$. Then the value of $a$ is ...........
A long, straight wire is surrounded by a hollow, thin, long metal cylinder whose axis coincides with that of wire. The wire has a charge per unit length of $\lambda$, and the cylinder has a net charge per unit length of $2\lambda$. Radius of the cylinder is $R$
If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero ? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.
$(a)$ Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
$\left( E _{2}- E _{1}\right) \cdot \hat{ n }=\frac{\sigma}{\varepsilon_{0}}$
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 $\sigma \hat{ n } / \varepsilon_{0}$
$(b)$ Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
Consider a metal sphere of radius $R$ that is cut in two parts along a plane whose minimum distance from the sphere's centre is $h$. Sphere is uniformly charged by a total electric charge $Q$. The minimum force necessary to hold the two parts of the sphere together, is