Question

The plates $\text{A}$ and $\text{B}$ of a parallel plate capacitor (PPC) as shown in the figure contain charges $\text{+Q}$ and $\text{-Q}$ respectively. The plate $\text{A}$ is fixed and the plate $\text{B}$ is connected to one end of a spring of spring constant $K$. The other end of the spring is connected to a rigid support. If the system is released from rest, what is the maximum elongation produced in the spring? The area of each plate is $\text{A}$ and the medium between the plates is air.

• A. $\frac{Q^2}{2KA{\epsilon }_o}$
• B. $\frac{Q^2}{KA{\epsilon }_o}$
• C. $\frac{2Q^2}{KA{\epsilon }_o}$
• D. $\frac{Q^2}{4KA{\epsilon }_o}$

Answer 1

The correct option is B $\frac{Q^2}{KA{\epsilon }_o}$


Plate $\text{B}$ starts from rest initially.

Lets assume that it moves by ${}^{\prime }{x}^{\prime }$ before it comes to rest again.

According to Work-Energy Theorem,

$W_{all}=\Delta K.E$

The forces acting on the plate $\text{B}$ are electrostatic force due to charge on the plate $\text{A}$ and spring force.

$\Rightarrow W_e+W_s=K_f-K_i......\left(i\right)$

As we know, electrostatic force between the plates of a PPC is $\frac{Q^2}{2{\epsilon }_{o}A}$

Using $\left(i\right)$ we get,

$\therefore \frac{Q^2}{2{\epsilon }_{o}A}x-\frac{1}{2}K{x}^2=0$

$\Rightarrow x=\frac{Q^2}{K{\epsilon }_{o}A}$

Hence, option (b) is the correct answer.

Why this question ? To understand the combination of mechanics with electrostatics.