A charge \(q\) is placed in a uniform electric field \(E.\) If it is released, then the kinetic energy of the charge after travelling distance \(y\) will be:

In the Millikan oil drop experiment, a charged drop falls with a terminal velocity \(v.\) If an electric field \(E\) is applied vertically upwards it moves with terminal velocity \(2v\) in the upward direction. If the electric field reduces to \(\frac{E}{2}\) then its terminal velocity will be:
1. \(\frac{v}{2}\)
2. \(v\)
3. \(\frac{3v}{2}\)
4. \(2v\)

The electric field at centre \(O\) of a semicircle of radius \(a\) having linear charge density \(\lambda\) is given by: