Answer:
a) the magnitude of the force is
F= Q(
) and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E = 
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) =
, where p = q × s
E(r) =
where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q(
)
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Answer:
Explanation:
Let the thickness of the film is t and the refractive index of the material of film is n.
When light travels through a sheet of thickness t, the optical path traveled is nt.
When the path of one of slit is covered by a sheet of thickness t, the optical path becomes
x = ( n - 1) t
As the one fringe is shift, so the optical path changed by one wavelength.
i.e., x = λ
So, λ = ( n - 1) t


<h3>A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?</h3>

<h3>Initial Velocity (<em>u</em>) - 5 m/s</h3><h3>Final Velocity (<em>v</em>) - 15 m/s</h3><h3>Time (<em>t</em>) - 10 sec</h3>

<h3>If the velocity of an object changes from an initial value <em>u </em>to the final value <em>v </em>in time <em>t,</em><em> </em>the acceleration <em>a</em> is, </h3><h3>

</h3><h3>

</h3>


<h3>His acceleration is </h3><h3>

</h3><h3 /><h3 />
Answer:
The vector form is as shown in the attachment
Explanation:
The figure as shown in the diagram, indicates that the car is moving along the road at a constant speed. Centripetal acceleration comes into play for an object moving in a circular motion at uniform speed. The centripetal acceleration is the acceleration experienced by an object while in uniform circular motion.
Mathematically from centripetal acceleration; a = v2/r
The equation shows that there is an inverse relationship between the acceleration and the radius of curvature as such the radius of curvature at the point A will be more than the radius of curvature at the point C, this shows that the centripetal acceleration at point C will be more than the centripetal acceleration at point A.
The attachment shows the figure and the representation in vectorial form.