Question

One plate of a capacitor is fixed, and the other is connected to the spring as shown in the figure. The area of both plates is A. In the steady-state (equilibrium), the separation between the plates is 0.8d (spring has unstretched). The force constant of the spring is approximately

a)\[\dfrac{125{{\varepsilon }_{0}}A{{E}^{2}}}{32{{d}^{3}}}\]
b) \[\dfrac{2{{\varepsilon }_{0}}A{{E}^{2}}}{{{d}^{3}}}\]
c) \[\dfrac{6{{\varepsilon }_{0}}{{E}^{2}}}{A{{d}^{3}}}\]
d) \[\dfrac{{{\varepsilon }_{0}}A{{E}^{2}}}{{{d}^{3}}}\]

Answer 1

Hint: Since we need to find the force constant of spring, so we need the value of force on the spring that is equal to the force on plates due to charged capacitor, i.e. $F=\dfrac{{{Q}^{2}}}{2A{{\varepsilon }_{0}}}$
Find the value of Q by the formula: $Q=CE;C=\dfrac{{{\varepsilon }_{0}}A}{D}$, where D = 0.8d
Then, by using the formula: $F=kx$, find the value of k for x = 0.2d

Complete step by step answer:
As we know that:
$Q=CE$, where Q is a charge, C is the capacitance and E is the magnitude of the electric field.
So, substituting the value of C in the above equation, we get:
$\begin{align}
  & Q=\dfrac{{{\varepsilon }_{0}}A}{D}E \\
 & =\dfrac{{{\varepsilon }_{0}}A}{0.8d}E......(1)
\end{align}$
Now, using the formula: $F=\dfrac{{{Q}^{2}}}{2A{{\varepsilon }_{0}}}$, find the value of force on plates due to charged capacitor, i.e.:
$\begin{align}
  & F=\dfrac{{{\left( \dfrac{{{\varepsilon }_{0}}A}{0.8d}E \right)}^{2}}}{2A{{\varepsilon }_{0}}} \\
 & =\dfrac{100{{\varepsilon }_{0}}A{{E}^{2}}}{128{{d}^{2}}}......(2)
\end{align}$
As we know that force on the spring that is equal to the force on plates due to charged capacitor
Therefore, using the formula: $F=kx$ for x = 0.2d, we get:
$\begin{align}
  & \Rightarrow \dfrac{100{{\varepsilon }_{0}}A{{E}^{2}}}{128{{d}^{2}}}=k\left( 0.2d \right) \\
 & \Rightarrow k=\dfrac{1000{{\varepsilon }_{0}}A{{E}^{2}}}{256{{d}^{3}}} \\
 & \Rightarrow k=\dfrac{125{{\varepsilon }_{0}}A{{E}^{2}}}{32{{d}^{3}}} \\
\end{align}$

So, the correct answer is “Option A”.

Note:
Derivation of force between two plates of the capacitor:
As we know that, the magnitude of the electric field by anyone plate is: $E=\dfrac{\sigma }{2{{\varepsilon }_{0}}}=\dfrac{Q}{2A{{\varepsilon }_{0}}}$, where A is the area of the plate.
Also, we know that: $\left| F \right|=\left| Q \right|\left| E \right|$
So, we have a force between two plates of the capacitor as:
$F=\dfrac{{{Q}^{2}}}{2A{{\varepsilon }_{0}}}$