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 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 upward direction. If 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: