Q.A capacitor is made of two circular plates of radius $R$ each, separated by a distance $d \ll R$. The capacitor is connected to a constant voltage. A thin conducting disc of radius $r \ll R$ and thickness $t \ll r$ is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is $m$.

Concept and Intuition

The energy density stored in the electric field of a parallel plate capacitor is a local property of the field strength, $u = \tfrac12\varepsilon_0 E^2$. It does not depend on the plate area or capacitance directly — only on how strong the field is, which is set by $E = V/d$. This is a useful distinction from total energy $U=\tfrac12CV^2$, which does depend on the geometry (through $C$). Since the question only asks for energy density, the plate area/capacitance is redundant information here.

Step-by-Step Solution

1. Convert the plate separation: $d = \dfrac{3}{\sqrt{\pi}}\ mm = \dfrac{3\times10^{-3}}{\sqrt{\pi}}\ m$.
2. Electric field: $E = \dfrac{V}{d} = \dfrac{180}{3\times10^{-3}/\sqrt{\pi}} = \dfrac{180\sqrt{\pi}}{3\times10^{-3}} = 6\times10^{4}\sqrt{\pi}\ V/m$.
3. $E^2 = 36\times10^{8}\times\pi = 3.6\times10^9\pi\ (V/m)^2$.
4. Energy density: $u = \tfrac12\varepsilon_0 E^2 = \tfrac12 \times 8.85\times10^{-12}\times 3.6\times10^9\pi$. $= 0.5 \times 8.85\times3.6\times\pi \times 10^{-3} = 0.5\times31.86\times3.1416\times10^{-3}$ $\approx 0.5\times100.1\times10^{-3} \approx 50.0\times10^{-3}\ J/m^3$. …

Concept and Intuition

For a capacitor held at constant voltage by a battery, the force between the plates can be found from how the stored energy changes as plate separation $t$ changes, using $F = \tfrac12 V^2\,\dfrac{dC}{dt}$ (the factor of $\tfrac12$ survives because although the battery supplies energy as $t$ changes, exactly half of the battery's work goes into mechanical work and half into the field energy change — the standard result for constant-voltage capacitor systems).

Step-by-Step Solution

1. $C = \dfrac{\varepsilon_0 A K}{t}$.
2. $\dfrac{dC}{dt} = -\dfrac{\varepsilon_0 A K}{t^2}$ (magnitude $\dfrac{\varepsilon_0 AK}{t^2}$).
3. Force magnitude: $F = \dfrac12 V^2 \left|\dfrac{dC}{dt}\right| = \dfrac12 \dfrac{\varepsilon_0 K A V^2}{t^2}$. …

Concept and Intuition

The energy stored in a charged capacitor is $W=\dfrac{Q^2}{2C}=\dfrac{1}{2}CV^2=\dfrac{1}{2}QV$. Of these forms, $W=\dfrac{Q^2}{2C}$ is most convenient here because the capacitance $C$ is fixed by the capacitor's geometry and does not change as we push more charge onto the plates. Since $W\propto Q^2$, energy does not scale linearly with charge — it scales quadratically. This is the key intuitive point: pushing the second increment of charge onto the plates costs more work than the first, because the plates are already at a higher potential opposing the new charge.

Step-by-Step Solution

1. Let the original charge be $Q$, so the original stored energy is $W = \dfrac{Q^2}{2C}$.
2. When the charge is doubled to $2Q$, the new stored energy is

3. The additional work required to take the capacitor from charge $Q$ to charge $2Q$ is the increase in stored energy: …

Concept and Intuition

A charged capacitor stores energy in its electric field. This energy can be expressed in three equivalent forms — $U = \tfrac12CV^2 = \tfrac12QV = \dfrac{Q^2}{2C}$ — and since here the charge and capacitance are given directly (not the voltage), the $Q^2/2C$ form is the most direct to use.

Step-by-Step Solution

1. Given: $C = 15\ \mu\text{F} = 15\times10^{-6}$ F, $Q = 300\ \mu\text{C} = 300\times10^{-6}$ C.
2. Energy stored: $U = \dfrac{Q^2}{2C} = \dfrac{(300\times10^{-6})^2}{2\times15\times10^{-6}}$.
3. Numerator: $(300\times10^{-6})^2 = 9\times10^{-8}$.
4. Denominator: $2\times15\times10^{-6} = 3\times10^{-5}$. …

Concept and Intuition

When two capacitors are joined plate-to-plate of the same polarity, charge conservation fixes their common final potential to the charge-weighted average — here, since both have the same capacitance $C$, the common potential is simply the arithmetic mean $V=\tfrac{V_1+V_2}{2}$. But energy is not conserved during this sudden redistribution (some is lost to resistive/radiative dissipation in the connecting wires), so the final stored energy is always less than the initial stored energy unless $V_1=V_2$.

Step-by-Step Solution

1. Initial energy: $U_i = \dfrac{1}{2}CV_1^2 + \dfrac{1}{2}CV_2^2 = \dfrac{C}{2}(V_1^2+V_2^2)$.
2. Total charge conserved, common potential: $V = \dfrac{CV_1+CV_2}{2C} = \dfrac{V_1+V_2}{2}$. …

Concept and Intuition

Charging a capacitor stores energy in its electric field, given by $U = \frac12 CV^2$ (equivalently $\frac12 QV$ or $\frac{Q^2}{2C}$). Because energy depends on $V^2$, doubling the voltage quadruples the stored energy — this quadratic sensitivity is why high-voltage capacitor banks store so much energy even with modest capacitance.

Step-by-Step Solution

1. Given $C = 10\,\mu\mathrm F = 10\times10^{-6}\,\mathrm F$ and $V = 6\,\mathrm{kV} = 6000\,\mathrm V$.
2. Apply $U=\dfrac12 C V^2$.
3. $V^2 = (6000)^2 = 3.6\times10^7\ \mathrm{V^2}$. …

Concept and Intuition

For capacitors in series, the equivalent capacitance is smaller than either individual value (they share the same charge, not the same voltage), given by $\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$. Once the equivalent capacitance across the full battery voltage is known, the energy stored in the whole combination is simply $U=\dfrac12 C_{eq}V^2$, since energy storage only cares about the total capacitance seen by the source and the voltage across it.

Step-by-Step Solution

1. Equivalent series capacitance: $C_{eq}=\dfrac{C_1 C_2}{C_1+C_2}=\dfrac{4\times6}{4+6}=\dfrac{24}{10}=2.4\ \mu F = 2.4\times10^{-6}$ F.
2. Energy stored: $U=\dfrac12 C_{eq}V^2=\dfrac12\times2.4\times10^{-6}\times(200)^2$. …

Concept and Intuition

A charge fully enclosed by a conducting shell and then connected to it by a wire cannot remain on the inner conductor in electrostatic equilibrium — any residual charge there would create a field inside the shell material, which is impossible for a conductor at equilibrium. So all the charge relocates to the outermost surface. The field that used to exist between $R$ and $2R$ disappears, and the energy once stored in that field is dissipated as heat in the wire (resistive dissipation during the transient charge flow).

Step-by-Step Solution

1. Before connection: field exists everywhere from $r=R$ outward (charge $Q$ effectively acts as if concentrated at the center for $r>R$, since the shell is neutral overall). Field energy $U_i = \dfrac{Q^2}{8\pi\varepsilon_0 R}$ (the standard energy of a point/sphere charge's field down to radius $R$: $U = \int_R^\infty \frac{1}{2}\varepsilon_0 E^2\,4\pi r^2\,dr = \frac{Q^2}{8\pi\varepsilon_0 R}$). …

Concept and Intuition

When plate separation doubles, capacitance halves ($C=\varepsilon_0A/d$). What stays constant depends on whether the battery is connected: disconnected means charge $Q$ is conserved (isolated system); connected means voltage $V$ is fixed by the battery. Energy formulas differ depending on which variable ($Q$ or $V$) is held fixed, which is exactly why $E_1$ and $E_2$ come out so different.

Step-by-Step Solution

1. Initial state: capacitance $C_0$, voltage $V_0$, charge $Q_0=C_0V_0$, energy $=\tfrac12 C_0V_0^2$.
2. Case 1 (battery disconnected): separation doubles, so new capacitance $C'=C_0/2$. Charge is conserved (no path for charge to change): $Q=Q_0$. Energy $E_1 = \dfrac{Q_0^2}{2C'} = \dfrac{(C_0V_0)^2}{2(C_0/2)} = \dfrac{C_0^2V_0^2}{C_0} = C_0V_0^2$.
3. Case 2 (battery connected): separation doubles, capacitance again $C_0/2$, but now voltage is held fixed at $V_0$ by the battery. …

Concept and Intuition

When capacitors are in series with a battery, they carry the same charge magnitude $Q$ (charge conservation along the single conducting loop), even though they have different capacitances and hence different voltages. When the battery is removed and the two capacitors are then reconnected directly to each other with their polarities reversed relative to the original loop (plate of one polarity joined to the opposite-polarity plate of the other), the charges that meet at each junction are equal in magnitude but of opposite sign — they cancel completely, since $Q$ was the same on both.

Step-by-Step Solution

1. Series equivalent capacitance: $C_{eq}=\dfrac{C\cdot2C}{C+2C}=\dfrac{2C}{3}$.
2. Charge on each capacitor (same in series): $Q = C_{eq}E = \dfrac{2CE}{3}$.
3. Voltages: $V_C = \dfrac{Q}{C}=\dfrac{2E}{3}$, $V_{2C}=\dfrac{Q}{2C}=\dfrac{E}{3}$.
4. Initial stored energy: $U_i = \dfrac{Q^2}{2C}+\dfrac{Q^2}{2(2C)} = \dfrac{Q^2}{2C}+\dfrac{Q^2}{4C}=\dfrac{3Q^2}{4C}$. Substituting $Q=\dfrac{2CE}{3}$: $U_i = \dfrac{3}{4C}\times\dfrac{4C^2E^2}{9}=\dfrac{CE^2}{3}$.
5. Reconnecting with opposite polarity together: at each junction the charges are $+Q$ from one capacitor and $-Q$ from the other (since the polarities are now reversed relative to each other) — net charge at each node is $Q-Q=0$. …

Concept and Intuition

Building up charge on a shell requires bringing successive infinitesimal charges from infinity against the repulsion of charge already placed. Integrating this work over the whole process (equivalently, integrating the field energy density $\tfrac12\varepsilon_0E^2$ over all space outside the shell, where $E=kQ/r^2$ for $r>R$ and $E=0$ inside) gives the shell's total electrostatic self-energy.

Step-by-Step Solution

1. Field outside a uniformly charged shell (charge $q$ already present) is $E=\dfrac{kq}{r^2}$; field energy density $u=\tfrac12\varepsilon_0E^2$. …

Concept and Intuition

Charge is conserved when two capacitors are joined, but the total electrostatic energy is not. As charge flows from the charged capacitor into the uncharged one, a transient current passes through the (small but non-zero) resistance of the wires, and part of the stored energy is irreversibly converted to heat. For two equal capacitors, the charged one ends up sharing its charge equally, the common voltage falls to half, and exactly half of the original energy is lost.

Step-by-Step Solution

1. Let $C = 50\ \text{pF} = 50\times10^{-12}\ \text{F}$ and $V = 100\ \text{V}$. Charge stored: $Q = CV = 5\times10^{-9}\ \text{C}$.
2. Initial energy: $U_i = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(50\times10^{-12})(100)^2 = 2.5\times10^{-7}\ \text{J} = 250\ \text{nJ}$.
3. After connecting to an identical uncharged capacitor, total capacitance $= 2C$; charge $Q$ is unchanged, so common potential $V' = \dfrac{Q}{2C} = \dfrac{5\times10^{-9}}{100\times10^{-12}} = 50\ \text{V}$.
4. Final energy: $U_f = \tfrac{1}{2}(2C)V'^2 = \tfrac{1}{2}(100\times10^{-12})(50)^2 = 1.25\times10^{-7}\ \text{J} = 125\ \text{nJ}$.
5. Energy lost: $\Delta U = U_i - U_f = 250 - 125 = 125\ \text{nJ} = 0.125\ \mu\text{J}$. Equivalently, $\Delta U = \tfrac{1}{2}\dfrac{C_1 C_2}{C_1 + C_2}V^2 = \tfrac{1}{2}(25\times10^{-12})(100)^2 = 0.125\ \mu\text{J}$.

Common Mistakes …