Answer:
A.)1.52cm
B.)1.18cm
Explanation:
angular speed of 120 rev/min.
cross sectional area=0.14cm²
mass=12kg
F=120±12ω²r
=120±12(120×2π/60)^2 ×0.50
=828N or 1068N
To calculate the elongation of the wire for lowest and highest point
δ=F/A
= 1068/0.5
δ=2136MPa
'E' which is the modulus of elasticity for alluminium is 70000MPa
δ=ξl=φl/E =2136×50/70000=1.52cm
δ=F/A=828/0.5
=1656MPa
δ=ξl=φl/E
=1656×50/70000=1.18cm
Answer:
a) E = 8.63 10³ N /C, E = 7.49 10³ N/C
b) E= 0 N/C, E = 7.49 10³ N/C
Explanation:
a) For this exercise we can use Gauss's law
Ф = ∫ E. dA = /ε₀
We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product
The area of a sphere is
A = 4π r²
if we use the concept of density
ρ = q_{int} / V
q_{int} = ρ V
the volume of the sphere is
V = 4/3 π r³
we substitute
E 4π r² = ρ (4/3 π r³) /ε₀
E = ρ r / 3ε₀
the density is
ρ = Q / V
V = 4/3 π a³
E = Q 3 / (4π a³) r / 3ε₀
k = 1 / 4π ε₀
E = k Q r / a³
let's calculate
for r = 4.00cm = 0.04m
E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³
E = 8.63 10³ N / c
for r = 6.00 cm
in this case the gaussine surface is outside the sphere, so all the charge is inside
E (4π r²) = Q /ε₀
E = k q / r²
let's calculate
E = 8.99 10⁹ 3 10⁻⁹ / 0.06²
E = 7.49 10³ N/C
b) We repeat in calculation for a conducting sphere.
For r = 4 cm
In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.
E = 0
In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is
E = k q / r²
E = 7.49 10³ N / C
Answer:
Electric field,
Explanation:
It is given that,
Magnitude of charge,
Force experienced,
We need to find the electric field at the origin. It is given by :
So, the electric field at the origin is . Hence, this is the required solution.