Nootan Isc Physics Class 12 Pdf 281 LINK ⭐

by idarkal · 22nd June 2022

Nootan Isc Physics Class 12 Pdf 281

In this section we will study the general behaviors of physical systems in a relatively ﬁxed
atmosphere, i.e. we assume the earth, and therefore the ball, to be at rest in the center of the
universe. Mass-energy is conserved in all events of this section, therefore, F = m1 + m2 = m1 M = (m1 m2 /
1/2) – mc2 is conserved in all the physical processes discussed here, and because this is not
just any electric field that is involved but a physical field, there is a physical consequence
involved.
In the limit, all energies are very small, $ \leq \, mc2$
, and the non-zero particle masses are also very small. This is typical of a very short range
electrostatic field.
5 Using this set of equations, calculate the minimum and maximum values of the electric
field that can be maintained by a charged particle moving in a circular path in an inﬂuential
electrostatic field.
Solution:
5
i.
ii.
iii.
iv.
v.
Use the differential equation approach: $ F = \Delta \,\overline{E}$
where $\Delta$ is the difference between the maximum and the minimum values of the
field. Therefore, $ \Delta \,\overline{E} = \frac{q \,\Delta \,v}{2 \,\pi \,r} $
Using the initial assumptions, we have $ \overline{E} \,\,\ll \, mc2 $
Therefore, $ F = \Delta \,\overline{E} \approx \frac{q \,\Delta \,v}{2 \,\pi \,r} $
The minimum value of $ \Delta \,v $ is when $ r=R $, the radius of the circle. $ \Delta \,v = R \,\Delta \,\theta $
The maximum value of $ \Delta \,v $ is when $ r=0 $, the center of the circle. $ \Delta \,v = \Delta \,\theta $
$ \Delta \,\overline{E} = \frac{q \,\Delta \,v}{2 \,\pi \,r} = \frac{