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JEE Main
Exam
JEE Main
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Physics
Chapter
Capacitance

JEE Main · Physics · Capacitance

One plate of a parallel plate capacitor is connected to a spring as...

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 AA and the distance between the plates is dd, 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 0.75d0.75 d, then the force constant of the spring is One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the

Options

  1. A38ε0V2Ad3\frac{3}{8} \frac{\varepsilon_0 V^2 A}{d^3}
  2. B83ε0V2Ad3\frac{8}{3} \frac{\varepsilon_0 V^2 A}{d^3}
  3. C932ε0V2Ad3\frac{9}{32} \frac{\varepsilon_0 V^2 A}{d^3}
  4. D329ε0V2Ad3\frac{32}{9} \frac{\varepsilon_0 V^2 A}{d^3}

Correct answer

D. 329ε0V2Ad3\frac{32}{9} \frac{\varepsilon_0 V^2 A}{d^3}

Step-by-step solution

In equilibrium, force between plates of capacitor = spring force ⇒q22ε0A=kx\Rightarrow \frac{q^2}{2 \varepsilon_0 A}=k x where, x=x= extension in spring. Now before charging, when spring is unstretched (x=0)(x=0) distance of plates is dd and after charging it is 34d\frac{3}{4} d. So, x=d−34d=14d\quad x=d-\frac{3}{4} d=\frac{1}{4} d Hence, k=q22ε0A⋅x=C2V22ε0Ax={ε02A2(34d)⋅V22ε0A⋅14d}=329(ε0V2Ad3)\begin{aligned} k & =\frac{q^2}{2 \varepsilon_0 A \cdot x}=\frac{C^2 V^2}{2 \varepsilon_0 A x} \\ & =\left\{\frac{\frac{\varepsilon_0^2 A^2}{\left(\frac{3}{4} d\right)} \cdot V^2}{2 \varepsilon_0 A \cdot \frac{1}{4} d}\right\}=\frac{32}{9}\left(\frac{\varepsilon_0 V^2 A}{d^3}\right)\end{aligned}
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