Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer:
116.3 electrons
Explanation:
Data provided in the question:
Time, t = 2.55 ps = 2.55 × 10⁻¹² s
Current, i = 7.3 μA = 7.3 × 10⁻⁶ A
Now,
we know,
Charge, Q = it
thus,
Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)
or
Q = 18.615 × 10⁻¹⁸ C
Also,
We know
Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C
Therefore,
Number of electrons past a fixed point = Q ÷ q
= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]
= 116.3 electrons
Answer:
Q=339.5W
T2=805.3K
Explanation:
Hi!
To solve this problem follow the steps below, the procedure is attached in an image
1. Draw the complete outline of the problem.
2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.
3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature
