imagine you are outside enjoying the warm sunshine with friends. as you briefly glance up at the sun, the part of the sun that y
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Answer:
a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2
b) See the picture
Explanation:
We can use Gauss's law to find the electric field in all the regions:
EA = qen/e0 where qen is the enclosed charge
Remember that the electric field everywhere outside a sphere is:
E(r) = q/(4*pi*eo*r^2) = Kq/r^2
a)
- For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
- For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
- For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2
b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance
Answer:
KE = 9.6MeV
Explanation:
Given the relationships we understand that,
The Potential at point p, is given by the following formula
According to the graphic designed, you have,
The kinetic energy of the particle would be given by
