A water droplet is placed in the field of a point charge of `10^(-5) C`. How far from the droplet must the charge be for the electric forces to overcome the force of gravity? The
In all problemos involving plane capacitors fringe effects ar to bo ignorod. radius of the droplet is much less than the distance between the droplet and the charge.

Explanation

The droplet is polarized in the electric field and acquires a dipole moment `p_(e) = PV`. where `P=chi_(e) epsi_(0)E` is the magnitude of the polarization vector and V is the volume of the droplet. In a non-uniform field a force `F=p_(e)=(dE)/(dr)` acts on the droplet. We have
`F=chi_(e) epsi_(0)VE=(dE)/(dr)=(epsi-1) epsi_(0)V(Q)/(4pi epsi_(0)r^(2)) (-2Q)/(4pi epsi_(0)r^(3))=-((epsi-1)VQ^(2))/(8 pi^(2) epsi_(0)r^(5)`
This force is directed upwards (Fig. 25.6). The force of gravity `mg= rho Vg` acts downwards. The problem require that the magnitude of the electric force exceed the magnitude of the force of gravity:
`((epsi-1)VQ^(2))/(8 pi^(2) epsi_(0) r^(5)) ge rho V, ` from Which `r le ""^(5) sqrt(((epsi-1)Q^(2))/(8pi^(2)epsi_(0)rhog))`