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

The electric potential $V(x)$ in a region around the origin is given by $V(x) = 4x^2 \text{ V}$. The electric charge enclosed in a cube of $1 \text{ m}$ side with its centre at the origin is (in coulomb):

Two large parallel plates have uniform surface charge densities $+\sigma$ and $-\sigma$. The electric field between the plates is:

$A$ charged ball $B$ hangs from a silk thread $S$,which makes an angle $\theta$ with a large charged conducting sheet $P$,as shown in the figure. The surface charge density $\sigma$ of the sheet is proportional to

$A$ non-conducting solid sphere of radius $R$ has a uniform volume charge density $\rho$. The electric potential at the center of the sphere is related to the potential at the surface and outside the sphere due to this uniform charge distribution.
Statement-$1$: When a charge $q$ is moved from the surface to the center of the sphere,the change in its potential energy is $q\rho R^2 / 6\varepsilon_0$.
Statement-$2$: The electric field at a distance $r$ $(r < R)$ from the center of the sphere is $\rho r / 3\varepsilon_0$.

$A$ non-conducting solid sphere has radius $R$ and uniform charge density. $A$ spherical cavity of radius $\frac{R}{4}$ is hollowed out of the sphere. The distance between the center of the sphere and the center of the cavity is $\frac{R}{2}$. If the charge of the sphere is $Q$ after the creation of the cavity and the magnitude of the electric field at the center of the cavity is $E = K \left( \frac{Q}{4 \pi \epsilon_0 R^2} \right)$,determine the approximate value of $K$.