Answer:

Explanation:
The electric field inside a parallel plate capacitor is

where A is the area of one of the plates, and Q is the charge on the capacitor.
The electric force on the electron is

where q is the charge of the electron.
By definition the capacitance of the capacitor is given by

Plugging this identity into the force equation above gives

The work done by this force is equal to change in kinetic energy.
W = Fx = (30q)(0.05) = 1.5q = K
The charge of the electron is 
Therefore, the kinetic energy is 
Options:
A. The luminosity of Star 1 is a factor of 100 less than the luminosity of Star 2.
B. Star 1 is 100 times more distant than Star 2.
C. Without first knowing the distances to these stars, you cannot draw any conclusions about how their true luminosities compare to each other.
D. Star 1 is 10 times more distant than Star 2.
E. Star 1 is 100 times nearer than Star 2.
Answer:
D. Star 1 is 10 times more distant than star 2
Explanation:
For two stars of identical size and temperature, the closer one to us will appear brighter. The relationship between the distance and luminosity of stars is an inverse- square relationship.
Luminosity, L = 1/r²
Where r is the distance of the star to the earth
Since star 1 is dimmer in brightness than star 2 by a factor of 100,
L₁/L₂ = 1/100
i.e. L₁ = 1, L₂=100
L₁ = 1/r₁² ............(1)
1 = 1/r₁²
L₂ = 1/r₂²
100 = 1/r₂² .........(2)
divide equation (2) by equation (1)
100/1 = ( 1/r₂² )/ (1/r₁²)
100 = (r₁/r₂)²
r₁/r₂ = √100
r₁/r₂ = 10
r₁ = 10r₂
D. Teach the public energy conservation
Answer:
E = k*Q₁/R₁² V/m
V = k*Q₁/R₁ Volt
Explanation:
Given:
- Charge distributed on the sphere is Q₁
- The radius of sphere is R₁
- The electric potential at infinity is 0
Find:
What is the electric field at the surface of the sphere?E.
What is the electric potential at the surface of the sphere?V
Solution:
- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.
- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.
- If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by
F = k*Q₁/R₁²
- Then the electric field at that point is
E = F/1
E = k*Q₁/R₁² V/m
- The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.
- Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation
V = k*Q₁/R₁ Volt
Answer:
y_red / y_blue = 1.11
Explanation:
Let's use the constructor equation to find the image for each wavelength
1 /f = 1 /o + 1 /i
Where f is the focal length, or the distance to the object and i the distance to the image
Red light
1 / i = 1 / f - 1 / o
1 / i_red = 1 / f_red - 1 / o
1 / i_red = 1 / 19.57 - 1/30
1 / i_red = 1,776 10-2
i_red = 56.29 cm
Blue light
1 / i_blue = 1 / f_blue - 1 / o
1 / i_blue = 1 / 18.87 - 1/30
1 / i_blue = 1,966 10-2
i_blue = 50.863 cm
Now let's use the magnification ratio
m = y ’/ h = - i / o
y ’= - h i / o
Red Light
y_red ’= - 5 56.29 / 30
y_red ’= - 9.3816 cm
Light blue
y_blue ’= 5 50,863 / 30
y_blue ’= - 8.47716 cm
The ratio of the height of the two images is
y_red ’/ y_blue’ = 9.3816 / 8.47716
y_red / y_blue = 1,107
y_red / y_blue = 1.11