Q.A slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $\frac{3}{4}d$, where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

The key idea here is the relationship between capacitance and the dielectric constant. For a parallel-plate capacitor, the capacitance is given by

where $A$ is the plate area, $d$ is the separation, and $k$ is the dielectric constant of the material between the plates. The only thing changing when we swap the dielectric for air is $k$ — everything else (geometry) stays fixed.

So, capacitance is directly proportional to $k$:

That means if you know the capacitance for one dielectric, you can find it for another by scaling.

1. Identify the given data

   • With dielectric ($k = 2$): $C_{\text{dielectric}} = 20\,\mu\text{F}$
   • With air ($k_{\text{air}} \approx 1$): we want $C_{\text{air}}$.

2. Write the proportionality

   Since $C \propto k$, we have

3. Plug in the numbers

So

4. Interpret the result …

Step 1 — Get the resonance condition.

In a series LCR circuit the impedance is

Resonance is where $Z$ is minimum, i.e. where the reactances cancel:

Step 2 — Isolate the dependence on $C$ (the only quantity being varied).

Step 3 — Translate into the shape of the graph.

• $f_0$ decreases as $C$ increases (a bigger capacitor stores charge more readily, lowering the natural oscillation frequency).
• $\dfrac{df_0}{dC}=-\tfrac12 \cdot \dfrac{1}{2\pi\sqrt L}\,C^{-3/2}$ — negative, and its magnitude shrinks as $C$ grows, so the curve is steep at small $C$ and flattens out. …

Allowed field: $E=10\%\times10^{6}=10^{5}\,\text{V m}^{-1}$.

Minimum plate separation for a $2\,\text{kV}$ rating:

Area from the capacitance $C=\dfrac{K\varepsilon_0 A}{d}$: …

Step 1 — What the battery fixes.

As long as the capacitor remains connected across the battery, the plates are held at the battery's emf. The potential difference cannot change:

This is the crux. (Contrast with the other standard version of this problem, where the battery is disconnected first — then the charge $Q$ is what stays fixed and the voltage changes instead. Read which one the question states!)

Step 2 — What removing the dielectric does to $C$.

With a dielectric of constant $K>1$ filling the gap,

Remove the slab and it becomes an air-gap capacitor:

Since $K > 1$,

$C_1 = K C_2 > C_2.$ …

Concept: a capacitor still connected to the battery keeps its VOLTAGE fixed (charge changes with C); an isolated capacitor keeps its CHARGE fixed (voltage changes).

From the figure: the 12 V cell, C1 and C2 all hang between the same top wire and the bottom wire, and the key K sits in the bottom wire BETWEEN the foot of C1 and the foot of C2. So C1 is permanently across the cell, while C2 completes its circuit only through K.

Stage 1 - K closed, both fully charged. Each capacitor sees the full 12 V:

Q1 = C1 V = 1 uF x 12 V = 12 uC

Q2 = C2 V = 1 uF x 12 V = 12 uC

Stage 2 - K opened. C2's lower plate is now disconnected, so C2 is ISOLATED and its charge is trapped at 12 uC. C1 remains connected across the 12 V cell. …

The original capacitor: $C=\dfrac{\varepsilon_0 A}{d}$.

When a dielectric occupies half the area, the two halves act as capacitors in parallel (same plate separation $d$, each of area $A/2$): …

Step 1 — Identify what is held constant. This is the whole question.

There are two standard dielectric-insertion scenarios and they give opposite answers:

• Battery still connected ⇒ $V$ is fixed (the battery enforces it), and $Q=CV$ increases.
• Battery disconnected ⇒ the charge on the isolated plates has nowhere to go, so $Q$ is fixed.

Here the capacitor is explicitly disconnected before the glass slab goes in, so

Step 2 — Capacitance.

A dielectric of dielectric constant $K>1$ (glass, $K\approx 5$–$10$) polarises and sets up an internal field opposing the applied one. The capacitance rises:

Step 3 — Potential difference.

Physically: the polarised dielectric partially cancels the field between the plates, $E=E_0/K$, and $V=Ed$, so $V$ drops by the factor $K$. Starting from $V_0=2\ \mathrm{V}$, the new p.d. is $2/K$ volts.

Step 4 — Energy stored.

With $Q$ fixed, the convenient form of the energy is the one written in terms of $Q$ and $C$:

The "lost" energy is real: the slab is pulled in by the fringing field, so the field does positive work on it — energy leaves the capacitor. …

1. The concept — the metre bridge is a Wheatstone bridge.

The bridge wire AB (uniform, 1 m long) is tapped by a jockey at D. At the null point the galvanometer carries no current, so the Wheatstone condition holds:

where $R$ and $S$ are the resistances in the two gaps and $R_{AD}, R_{DB}$ are the resistances of the two segments of the bridge wire.

2. Write the wire's resistance explicitly.

For a uniform wire of resistivity $\rho$, radius $r$ and length $\ell$:

3. Substitute — and watch $r$ cancel.

The resistivity $\rho$ and the cross-sectional area $\pi r^2$ cancel identically, because both segments are cut from the same wire. The balance condition therefore reduces to a pure ratio of lengths.

4. Apply the change. …

The key to this question is recognising the deep analogy between mechanical and electrical oscillations. A block on a spring oscillates because the spring provides a restoring force proportional to displacement: $F = -k x$. In an LC circuit, the charge on the capacitor oscillates because the capacitor provides a restoring "force" (voltage) proportional to charge: $V = -Q/C$. The minus sign indicates that the voltage always pushes charge back toward the equilibrium (zero charge). So the constant of proportionality here is $1/C$, which plays the same role as $k$ in the mechanical system.

Let’s walk through the analogy step by step to see why the other options don’t fit.

1. Write the equations of motion side by side. For a mass $m$ on a spring of constant $k$:

For an LC circuit (with inductance $L$ and capacitance $C$):

The form is identical: the coefficient in front of $x$ (or $Q$) on the right-hand side is the "restoring coefficient."

2. Identify the analogous quantities.

   Comparing the two equations:

   • Mass $m$ is analogous to inductance $L$ (both store kinetic/magnetic energy and resist change in velocity/current).
   • Displacement $x$ is analogous to charge $Q$.
   • The spring constant $k$ is analogous to $1/C$.

   So the electrical equivalent of $k$ is $1/C$, not $C$ itself.

3. Check the other options to confirm.

   • Capacitive reactance $X_C = 1/(\omega C)$ depends on frequency $\omega$, so it’s not a constant of the circuit like $k$ is a constant of the spring. …

Step 1 — Capacitance with a partially-filled gap.

For a parallel-plate capacitor of plate separation $d$ containing a slab of thickness $t$ and dielectric constant $K$ (the rest being air):

The reason: the slab's thickness $t$ is removed from the air path and re-inserted as an electrically shorter length $t/K$, because the dielectric weakens the field inside it by the factor $K$.

Step 2 — Read off the "effective shortening".

Compare the denominator with the original air-only gap $d$:

So the slab makes the gap behave as if it were shorter by $t\left(1 - \dfrac{1}{K}\right)$, which is why the capacitance goes up when the slab is inserted.

Step 3 — Impose the restoring condition.

The plates are then moved apart by $\Delta d = 3.5 \times 10^{-3}\,\mathrm{m}$ to bring $C$ back to its original value. "Back to original" means the effective separation must once again equal $d$, so the increase must exactly cancel the shortening:

Step 4 — Substitute and solve. …