A charge Q is distributed over two concentric conducting thin spherical shells radii r and R (R > r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is :

A charged particle (mass m and charge q)
moves along X-axis with velocity V₀. When it
passes through the origin it enters a region having uniform electric field
$$\overrightarrow E = - E\widehat j$$ which extends upto x = d.
Equation of path of electron in the region x > d is

Consider a sphere of radius R which carries a uniform charge density $$\rho $$. If a sphere of radius $${{R \over 2}}$$ is carved out of it, as shown, the ratio $${{\left| {\overrightarrow {{E_A}} } \right|} \over {\left| {\overrightarrow {{E_B}} } \right|}}$$ of magnitude of electric field $${\overrightarrow {{E_A}} }$$ and $${\overrightarrow {{E_B}} }$$, respectively, at points A and B due to the remaining portion is :

An electric dipole of moment
$$\overrightarrow p = \left( { - \widehat i - 3\widehat j + 2\widehat k} \right) \times {10^{ - 29}} $$ C.m is
at the origin (0, 0, 0). The electric field due to this dipole at
$$\overrightarrow r = + \widehat i + 3\widehat j + 5\widehat k$$ (note that $$\overrightarrow r .\overrightarrow p = 0$$ ) is parallel to :