The change in potential energy of the proton is 5.6 x
Joule
<h3>
What is a Uniform Electric Field ?</h3>
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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Answer:
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Answer:
1.635×10^-3m
Explanation:
Young modulus is the ratio of the tensile stress of a material to its tensile strain.
Young modulus = Tensile stress/tensile strain
Tensile stress = Force/Area
Given force = 130N
Area = Πr² = Π×(1.55×10^-3)²
Area = 4.87×10^-6m²
Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²
Tensile strain = extension/original length
Tensile strain = e/3.9
Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²
2×10¹¹N/m² = 8.39×10^7/{e/3.9)}
2×10¹¹ = (8.39×10^7×3.9)/e
2×10¹¹e = 3.27×10^8
e = 3.27×10^8/2×10¹¹
e = 1.635×10^-3m
The stretch of the steel wire will be
1.635×10^-3m
Answer:
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