What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radiu
1 answer:
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Complete question:
A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).
Required:
What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.
Answer:
The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Explanation:
Given;
charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C
length of the conductor, L = 50 m
inner radius, r₁ = 1.304 mm
outer radius, r₂ = 9.249 mm
The magnitude of the electric field halfway between the two cylindrical conductors is given by;
Where;
λ is linear charge density or charge per unit length
r is the distance halfway between the two cylindrical conductors
The magnitude of the electric field is now given as;
Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Answer: d. 8.25 m/s
Explanation:
We are given that Current= 5 m/s in j direction
Velocity= 8 m/s i + 3 m/s j
Now, we have to find Jada's speed with respect to the water.
First we find Jada's velocity with respect to water
v= (8 i + 3 j) - (5 j)
v= 8i - 2 j
To find the speed, we take the magnitude of this velocity vector we have
|v|=
|v|= = 8.246 m/s
which comes out to be around = 8.25 m/s
So option d is correct.
The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by
where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:
Learn more about potential energy:
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