Answer:
E = 1,873 10³ N / C
Explanation:
For this exercise we can use Gauss's law
Ф = E. dA =
/ ε₀
Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.
The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.
The surface of a sphere is
A = 4π r²
E 4π r² = q_{int} /ε₀
The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is
q_{int} = q₁ + q₂
q_{int} = (530 - 200) 10⁻⁹
q_{int} = 330 10⁻⁹ C
The electric field is
E = 1 / 4πε₀ q_{int} / r²
k = 1 / 4πε₀
E = k q_{int}/ r²
Let's calculate
E = 8.99 10⁹ 330 10⁻⁹/ 1.32²
E = 1,873 10³ N / C
Using the law of conservation of momentum
m1u1+m2u2=m1v1+m2v2
Where m1 is mass of first object
m2 is mass of second object
u1 and u2 are initial velocities of object 1 and 2 respectively
v1 and v2 are final velocities of object 1 and 2 respectively
Here, they are moving as a system after collision. Thus they will posses same final velocity
m1u1 +m2u2=v(m1+m2)
Substituting values
600*4+0=v(600+400)
2400=v*1000
v=2.4 m/s
Now momentum of system
p=Mv
p=(600+400)*2.4
p=1000*2.4
Therefore p=2400 kg m/s
Hope this helps :)
The answer would be it will be longer than the 656.3 nm. The reduced mass of positronium is less than hydrogen so the photon energy will be a reduced amount of for positronium than for hydrogen. So this will mean that the wavelength will be lengthier.