A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V = 12 V between its plates. The charging battery is now disconnected and a porcelin plate with k = 7 is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of pJ. (Assume no friction)

Key Concepts

1. Capacitance ($C$) – ability of a system to store charge per unit voltage. For a parallel-plate capacitor with empty space, $C$ is given.
2. Energy stored in a capacitor
   • Fixed voltage (battery connected): $U=\tfrac12 C V^2$
   • Fixed charge (battery disconnected): $U=\frac{Q^2}{2C}$
3. Dielectric constant ($k$) – inserting a dielectric multiplies the capacitance by $k$: $C' = k C$
4. Conservation of charge – with the battery unplugged, the total charge $Q$ on the plates stays the same.
5. Energy difference → Work – Any decrease in electrostatic energy must appear as another form (here, mechanical energy of the dielectric slab).

Logical Chain to Solve

1. Compute initial charge using $Q = C V$ before the battery is detached.
2. Find initial electrostatic energy with $U_i = \tfrac12 C V^2$ since the voltage is still $V$.
3. After the dielectric fully enters, capacitance becomes $C' = k C$ but charge remains $Q$. So new energy $U_f = \frac{Q^2}{2 C'}.$
4. Energy converted to motion = $\Delta U = U_i - U_f$
5. Express the answer in pico-Joules (pJ) for neatness.

That’s all the physics involved – no hidden tricks!

🌟 What’s happening here?

Think of a capacitor as two metal plates that can hold electric charge, just like two slices of bread that can store butter in-between.

1. We first charge the plates with a battery – that is like spreading butter (electric charge) on the slices.
2. We then disconnect the battery – no more butter can come or go; the butter you already spread must stay.
3. Now we slide a porcelain sheet (dielectric) between the plates. This sheet makes it easier for the plates to hold charge, so the electric field energy inside gets smaller.
4. Where does the “lost” energy go? Since nothing else can take it (we said no friction), that energy turns into the porcelain sheet’s kinetic energy. The sheet speeds up, overshoots a bit, slows down, then comes back, and keeps oscillating – like a child on a swing.

The question asks: How much mechanical energy (in pico-Joules) does the sheet keep swinging with?

Step-by-Step Calculation

1. Given data
   $C = 14\text{ pF} = 14 \times 10^{-12}\,\text{F}$
   $V = 12\,\text{V}$
   $k = 7$

2. Initial charge (battery still connected) $Q = C V = (14\times10^{-12}) \times 12 = 168 \times 10^{-12}\,\text{C} = 168\,\text{pC}$

3. Initial energy (battery now detached but voltage is still $V$ at that instant)

   $U_i = \tfrac12 C V^2$

   $= \tfrac12 \times 14\times10^{-12} \times (12)^2$

   $= \tfrac12 \times 14 \times 10^{-12} \times 144$

   $= 1.008 \times 10^{-9}\,\text{J} = 1008\,\text{pJ}$

4. Capacitance after inserting porcelain slab

   $C' = k C = 7 \times 14\,\text{pF} = 98\,\text{pF}$

5. Final energy with constant charge

   $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2 k C} = \frac{U_i}{k}$

   $U_f = \frac{1008\,\text{pJ}}{7} = 144\,\text{pJ}$

6. Mechanical energy gained by the dielectric slab

   $\Delta U = U_i - U_f = 1008\,\text{pJ} - 144\,\text{pJ} = 864\,\text{pJ}$

[ \boxed{\text{Mechanical energy} = 864\ \text{pJ}} ]