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:
C) W = - 190 J
Explanation:
Notation
Wf = work done by the friction force (unknown)
Ff = force of the friction
d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m
Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V
C1= 3×10^-6 F
V1= 480v
C2= 4×10^-6 F
V2= 500v
(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V
Simplifying the above, we get:
( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.
Further simplified as:
3440 × 10^-6 = 7 × 10^-6 × V
Making V the subject
V = 491.43volts
Therefore the potential difference across each capacitor is 491.43v
Answer:
40°
Explanation:
when a ray light is incident on a surface it makes a 40° angle with the normal to the surface (reflecting surface ). this called angle of incidence.
And the angle made by reflected ray with the normal to the surface is called angle of reflection.
by the law of reflection angle of incidence is equal to the angle of reflection
hence the answer to the question is 40° ( because ∠i =∠r )
Answer:
Explanation:
When a charged particle moves in a magnetic field , a force acts on it perpendicular to its velocity.
Force = Bqv , B is magnetic field , q is charge of the particle and v is velocity
= .03 x 6 x 70
= 12.6 N
