Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is $\alpha$. The spheres are now immersed in a dielectric liquid of density $800 kg m ^{-3}$ and dielectric constant $21$ . If the angle between the strings remains the same after the immersion, then

$(A)$ electric force between the spheres remains unchanged

$(B)$ electric force between the spheres reduces

$(C)$ mass density of the spheres is $840 kg m ^{-3}$

$(D)$ the tension in the strings holding the spheres remains unchanged

Two point charges of $20\,\mu \,C$ and $80\,\mu \,C$ are $10\,cm$ apart. Where will the electric field strength be zero on the line joining the charges from $20\,\mu \,C$ charge......$m$

Select the correct statement : (Only force on a particle is due to electric field)

A drop of ${10^{ - 6}}\,kg$ water carries ${10^{ - 6}}\,C$ charge. What electric field should be applied to balance its weight (assume $g = 10\,m/{s^2}$)

The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass and charge of an electron respectively are $9.1 \times 10^{-31}\,kg$ and $1.6 \times$ $10^{-19}\,C$.)

The distance between the two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$ At what point on the line joining the two, the intensity will be zero