AP EAMCET · PHYSICS · Capacitance
One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the capacitor is \(A\) and the distance between the plates is \(d\), when the battery is not connected and the spring is unstretched. After connecting the battery, in the steady state the distance between the plates is \(0.75 d\), then the force constant of the spring is
- A \(\frac{3}{8} \frac{\varepsilon_0 V^2 A}{d^3}\)
- B \(\frac{8}{3} \frac{\varepsilon_0 V^2 A}{d^3}\)
- C \(\frac{9}{32} \frac{\varepsilon_0 V^2 A}{d^3}\)
- D \(\frac{32}{9} \frac{\varepsilon_0 V^2 A}{d^3}\)
Answer & Solution
Correct Answer
(D) \(\frac{32}{9} \frac{\varepsilon_0 V^2 A}{d^3}\)
Step-by-step Solution
Detailed explanation
In equilibrium, force between plates of capacitor \(=\) spring force \(\Rightarrow \frac{q^2}{2 \varepsilon_0 A}=k x\) where, \(x=\) extension in spring. Now before charging, when spring is unstretched \((x=0)\) distance of plates is \(d\) and after charging it is…
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