Question

One of the plates of a charged parallel plate capacitor is connected to a non conducting spring of stiffness k while the other plate is fixed. The other end of the spring is also fixed in equilibrium, the distance between the plates is d, which is twice the elevation in the spring. If the length of the spring is halved by cutting it, the distance between the plates in equilibrium will be (consider that in both the cases spring is in natural length if the capacitor is uncharged)

Answer 1

Now the force of attraction between plates of capacitor is F=Q²/2A€ where Q is charge on capacitor plates, and A is the area of plates of capacitor and € permittivity of medium. Thus the force does not depend on the distance between the plates. Thus for this problem F should be constant as Q or A does not change. Now spring force =Kx where x is the elongation. It is given 2x=d In equilibrium, F=Kx Now when spring is halved the spring constant (stiffness) becomes 2K. Now in equilibrium, F=2Kx1 where x1 is new elongation. Therefore x1=x/2=d/4. Now I don't see how I can get the new distance between plates as the natural length of spring is not mentioned.