ElectrostaticsHard
Question
An alpha particle of energy 5 MeV is scattered through 180o by a fixed uranium nucleus. The distance of closest approach is of the order :
Options
A.1(Ao)
B.10-10 cm
C.10-12 cm
D.10-15 cm
mv2 = 
⇒ d = 10-12 cmMore Electrostatics Questions
The magnitude of charges of electric dipole is 3.2 × 10-19 and distance between them is 2.4 . If it is placed in a ...If VA and VB are electric potentials at points A and B in given dig. then :-...Two identical thin rings, each of radius R meters, are coaxially placed at a distance R meters apart. If Q1 coulomb and ...15 joule of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. Then the pote...Plates A ard B constitutes an isolated, charge parallel plate capacitor. The inner surfaces (I and IV) of A and B have c...