Answer:
Explanation:
Let the plastic rod extends from - L to + L .
consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .
It will create a field at point P on y -axis . Distance of point P
= √ x² + .15²
electric field at P due to small charged length
dE = k λ dx x / (x² + .15² )
Its component along Y - axis
= dE cosθ where θ is angle between direction of field dE and y axis
= dE x .15 / √ x² + .15²
= k λ dx .15 / (x² + .15² )³/²
If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L
E = ∫ k λ .15 / (x² + .15² )³/² dx
= k λ x L / .15 √( L² / 4 + .15² )
This problem involves Newton's universal law of gravitation and the equation to follow would be.
F = GM₁M₂/r²
Given: M₁ = 0.890 Kg; M₂ = 0.890 Kg; F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²
Solving for distance r = ?
r = √GM₁M₂/F
r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N
r = 0.81 m
400m in 32sec: (400/32)>12.5meters per second>
(12.5)(60)(60)(1/1000)=45km per hour
Constant speed would mean that the two forces are equivalent