Question

Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant
‘k’. The charge on each capacitor is q. Find the extension in the spring at equilibrium

Answer 1

Answer:

it is a good question

Explanation:

force due to charges on capacitors is

F=kq1.q2/d^2

given charges are a

F=Kq^2/d^2

another force is present due to spring

F=kx^2

where x is elongation or extension

equating both

Ķq^2/d^2=kx^2

*(above k are different)

x^2=Ķq^2/d^2.k

x=under root(Ķq^2/d^2.k)

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Answer 2

Answer:

Hope this pic helps!!!