Answer:
The new Coulomb force is q₁q₂/9πε₀r²
Explanation
The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².
Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.
Since we have air between q₂ and q₃, the coulomb force between them is
F' = q₂q₃/4πε₀d²
= q₂(q₁/κ)/4πε₀(r/2)²
= 4q₂q₁/κ4πε₀r²
= 4/κ(q₂q₁/4πε₀r²)
= 4/9 × (q₂q₁/4πε₀r²)
= q₁q₂/9πε₀r²
So, the new Coulomb force is q₁q₂/9πε₀r²
Answer:
The tube should be held vertically and perpendicular to the ground.
Explanation:
Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:
Reasoning:
The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.
Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.
So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.
hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.
Answer:
D. The period would decrease by sqrt (2)
Explanation:
The period of a mass-spring system is given by:
where
m is the mass
k is the spring constant of the spring
If the spring constant is doubled,
k' = 2k
So the new period will be
So the correct answer is
D. The period would decrease by sqrt (2)
Answer:
The potential energy is 189nm
Explanation:
