What is one way to model gravity and magnetism
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Answer:
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
Explanation:
Energy conservation law: In isolated system the amount of total energy remains constant.
The types of energy are
- Kinetic energy.
- Potential energy.
Kinetic energy
Potential energy =
Here, q₁= +5.00×10⁻⁴C
q₂=-3.00×10⁻⁴C
d= distance = 4.00 m
V = velocity = 800 m/s
Total energy(E) =Kinetic energy+Potential energy
+
=(1280-337.5)J
=942.5 J
Total energy of a system remains constant.
Therefore,
E +
m/s
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
The change in potential energy of the proton is 5.6 x Joule
A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.
Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.
- The distance d = 10 cm = 0.1 m
- Electric field E = 3.5 KN/C
- Proton charge q = 1.6 x
C
The Work done = Fd
but F = Eq
Recall that Electric field E = F/q = V/d
Where V = potential difference.
Let us first calculate the V
E = V/d
V = Ed
Substitute all the parameters into the formula above
V = 3.5 × 10³ × 0.1
V = 350 v
from F/q = V/d
make F the subject of formula and substitute it in work formula
F = Vq/d
W.D = Vq/d x d
W.D = Vq
Substitute all the parameters into the formula above
W.D = 350 x 1.6 x
W.D = 5.6 x J
Work done = Energy = Potential Energy
Therefore, the change in potential energy of the proton is 5.6 x <em> Joule</em>
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