An electron travelling at 5 x 10^6cm s^-1 passes through an electric field that gives it an acceleration of 10^17 cm s^-2 acting in the direction of the travel, (a) How much time will it take for the electron to double its initial velocity? (b) How much distance will the electron travel in this time interval?
Answers
Answer:
Suppose we have a point charge q0 located at r and a set of external charges conspire so as
to exert a force F on this charge. We can define the electric field at the point r by:
E =
F
q0
(2.1)
The (vector) value of the E field depends only on the values and locations of the external
charges, because from Coulomb’s law the force on any “test charge” q0 is proportional to the
value of the charge. However to make this definition really kosher we have to stipulate that
the test charge q0 is “small”; otherwise its presence will significantly influence the locations
of the external charges.
Turning Eq. 2.1 around, we can say that if the electric field at some point r has the value
E then a small charge placed at r will experience a force
F = q0E (2.2)
The electric field is a vector . From Eq. 2.1 we can see that its SI units must be N
C
.
It follows from Coulomb’s law that the electric field at point r due to a charge q located
at the origin is given by
E = k
q
r
2
ˆr (2.3)
where ˆr is the unit vector which points in the same direction as r.
Answer:
This can be found out using first and second equations of motion.
Explanation: