A cube of side $b$ has a charge $q$ at each of its vertices. The electric field due to this charge distribution at the centre of this cube will be

Three charges $+Q, q, +Q$ are placed respectively, at distance, $0, \frac d2$ and $d$ from the origin, on the $x-$ axis. If the net force experienced by $+Q$, placed at $x = 0$, is zero, then value of $q$ is

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged is brought in contact with $B$, then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

Two charges $-\mathrm{q}$ each are fixed separated by distance $2\mathrm{d}$. A third charge $\mathrm{d}$ of mass $m$ placed at the midpoint is displaced slightly by $x (x \,<\,<\, d)$ perpendicular to the line joining the two fixed charged as shown in figure. Show that $\mathrm{q}$ will perform simple harmonic oscillation of time period.  $T =\left[\frac{8 \pi^{3} \epsilon_{0} m d^{3}}{q^{2}}\right]^{1 / 2}$

Positive charge $Q$ is distributed uniformly over a circular ring of radius $R$. A point particle having a mass $(m)$ and a negative charge $-q$ is placed on its axis at a distance $x$ from the centre. Assuming $x < R,$ find the time period of oscillation of the particle, if it is released from there [neglect gravity].