Question

A particle of mass M and charge q is released from rest in a region of uniform electric field magnitude E. After a time t, the distance travelled by the charge is S and the kinetic energy obtained by the particle is T . Then, the ratio T:S is
A.Remains constant with time
B.Varies linearly with the mass of the particle
C.Is independent of the charge
D.Is independent of the magnitude of the electric field

Answer 1

Hint: The least complex case happens when a charged molecule moves opposite to a uniform B-field. In the event that the field is in a vacuum, the electric field is the prevailing factor deciding the movement.

Complete answer:
The correct answer is A.
Work done by the field on the charge is= \[qES\]
\[\therefore T=qES\]
\[\because \dfrac{T}{S}=qE\]
Since the attractive power is opposite to the heading of movement, a charged molecule follows a bended path in an electric field. The molecule keeps on following this bended way until it frames a total circle. Another approach to see this is that attractive power is consistently opposite to speed, with the goal that it accomplishes no work on the charged molecule. The molecule's dynamic vitality and speed in this way stay steady. The bearing of movement is influenced however not the speed.
 The attractive power is opposite to the speed, so speed alters in course however not size. The outcome is uniform movement. (Note that in light of the fact that the charge is negative, the power is inverse in course to the forecast of the right-hand rule.)

Note:
An attractive power can gracefully centripetally power and cause a charged molecule to move in a roundabout way of radius=\[r=\dfrac{mv}{qB}\]
The time of round movement for a charged molecule moving in an attractive field opposite to the plane of movement is \[T=\dfrac{2\pi m}{qB}.\]
Helical movement results if the speed of the charged molecule has a segment corresponding to the attractive field just as a segment opposite to the attractive field.