Answer:
Explanation:
Given that,
The mass of a Hubble Space Telescope,
It orbits the Earth at an altitude of
We need to find the potential energy the telescope at this location. The formula for potential energy is given by :
Where
is the mass of Earth
Put all the values,
So, the potential energy of the telescope is .
Answer:
Solution is in explanation
Explanation:
part a)
For normalization we have
Part b)
Answer:
A)Magnitude of the net electric field at the origin due q₁ and q₂
E₀= 131.6 N/C
B) Direction of the net electric field at the origin due q₁ and q₂
β=3.91°
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻⁹ C
Data
q₁ = -4 nC = -4*10⁻⁹ C
q₂ = +6nC =+ 6*10⁻⁹ C
k = 9*10⁹ N*m²/C²
d₂ = 0.6 m
Graphic attached
The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:
E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.
E₂: Total field at point x=0 , y=0 due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.
Calculation of the electric field at the origin of x-y coordinates due to the charge q₁
E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C
Components (x-y) of the field due to q1:
E₁x=E₁*cosα = 36*(0.6/1) = 36*(0.6) = 21.6 N/C
E₁y=E₁*sinα = 36*(0.8/1) = 36*(0.8) =28.8 N/C
Calculation of the electric field at the origin of x-y coordinates due to the charge q₂
E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C
Calculation of the electric field components at the origin of x-y coordinates
E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)
E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C
E₀y= E₁y = 28.8 N/C
A )Magnitude of the net electric field at the origin due q₁ and q₂
E₀= 131.6 N/C
B) Direction of the net electric field at the origin due q₁ and q₂
β=3.91°
