Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:
DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation
according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero
solving for x we get
x = 0.200 m
b)electric field at half way mean x =0.25
E = 3.84*10^{-4} N/C
Answer:
<em>d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.</em>
<em></em>
Explanation:
Let us take the momentum of a photon unit as u
we know that the rate of change of momentum is proportional to the force exerted.
For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that
F = (u - 0)/t = u/t
for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,
F = u
For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to
F = (u - (-u))/t = 2u/t
just as the we did above, it becomes
F = 2u.
From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that <em>the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.</em>
<em></em>