Four charges are placed at a distance $a$ from the origin as shown in the figure. The dipole moment of the configuration is:

Two charges $+q$ and $-q$ are kept at a certain distance apart. Then, at any point on the right bisector of the line joining the two charges:

Three point charges $+q$,$-2q$ and $+q$ are placed at points $(x = 0, y = a, z = 0)$,$(x = 0, y = 0, z = 0)$ and $(x = a, y = 0, z = 0)$ respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

$A$ point $Q$ lies on the perpendicular bisector of an electric dipole with dipole moment $p$. If the distance of $Q$ from the dipole is $r$ (where $r$ is much larger than the size of the dipole),then the electric field at $Q$ is proportional to:

One end of a spring of negligible unstretched length and spring constant $k$ is fixed at the origin $(0,0)$. $A$ point particle of mass $m$ carrying a positive charge $q$ is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole $\overrightarrow{p}$ pointing towards the charge $q$ is fixed at the origin,the spring gets stretched to a length $\ell$ and attains a new equilibrium position. If the point mass is now displaced slightly by $\Delta \ell \ll \ell$ from its equilibrium position and released,it is found to oscillate at frequency $\frac{1}{\delta} \sqrt{\frac{k}{m}}$. The value of $\delta$ is. . . . . .

An electric dipole with dipole moment $\vec{p} = \frac{p_0}{\sqrt{2}}(\hat{i}+\hat{j})$ is held fixed at the origin $O$ in the presence of a uniform electric field $\vec{E} = E_0 \hat{i}$. If the potential is constant on a circle of radius $R$ centered at the origin as shown in the figure,then the correct statement$(s)$ is/are:
($\varepsilon_0$ is the permittivity of free space,$R \gg$ dipole size)
$(1)$ $R = \left(\frac{p_0}{4 \pi \varepsilon_0 E_0}\right)^{1/3}$
$(2)$ The magnitude of the total electric field on any two points of the circle will be the same.
$(3)$ The total electric field at point $A$ is $\vec{E}_A = \sqrt{2} E_0(\hat{i}+\hat{j})$
$(4)$ The total electric field at point $B$ is $\vec{E}_B = 0$