99 Percentile Qs Bank for JEE MainPhysicsCapacitance
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
Options
- A3 8 ₀ V^2 A d^3
- B8 3 ₀ V^2 A d^3
- C9 32 ₀ V^2 A d^3
- D32 9 ₀ V^2 A d^3
Correct answer
D. 32 9 ₀ V^2 A d^3
Step-by-step solution
In equilibrium, force between plates of capacitor = spring force q^2 2 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 3 4 d . So, x=d- 3 4 d= 1 4 d Hence, aligned k & = q^2 2 ₀ A x = C^2 V^2 2 ₀ A x & = ₀^2 A^2 ( 3 4 d ) V^2 2 ₀ A 1 4 d = 32 9 ( ₀ V^2 A d^3 ) aligned