The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is
1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,
The distance from the center of the square to one of the corners is
The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.
2)
Answer:
a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
Explanation:
Given that;
height of the ramp h1 = 0.40 m
foot of the ramp above the floor h2 = 1.50 m
assuming R = 15 mm = 0.015 m
density of steel = 7.8 g/cm³
density of aluminum = 2.7 g/cm³
a) distance that the solid steel sphere sliding down the ramp without friction;
we know that
distance = speed × time
d = vt --------let this be equ 1
according to the law of conservation of energy
mgh₁ = mv²
v² = 2gh₁
v = √(2gh₁)
from the second equation; s = ut + at²
that is; t = √(2h₂/g)
so we substitute for equations into equation 1
d = √(2gh₁) × √(2h₂/g)
d = √(2gh₁) × √(2h₂/g)
d = 2√( h₁h₂ )
we plug in our values
d = 2√( 0.40 × 1.5 )
d = 1.55 m
Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b)
distance that a solid steel sphere rolling down the ramp without slipping;
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c)
distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1 again
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.549√( h₁h₂ )
d = 1.549√( 0.4 × 1.5 )
d = 1.2 m
Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) distance that a solid aluminum sphere rolling down the ramp without slipping.
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m