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 $m$.

The key idea is that for a damped harmonic oscillator (like a mass on a spring with friction), the mechanical energy is proportional to the square of the amplitude. This is because energy in a simple harmonic oscillator is either all kinetic at the equilibrium point or all potential at the turning points, and both scale with $A^2$. Damping reduces the amplitude over time, so the energy drops even faster.

Why this approach works:

We don’t need the exact damping equation — just the relationship between amplitude and energy. If the amplitude halves, the energy becomes $(1/2)^2 = 1/4$ of the original. The percentage decrease is then $100\% - 25\% = 75\%$.

1. Recall the energy-amplitude relation For any simple harmonic oscillator (undamped or damped, at a given instant), the total mechanical energy is

where $k$ is the effective spring constant and $A$ is the instantaneous amplitude. This holds because at maximum displacement, all energy is potential.

2. Apply the given condition At time $t$, the amplitude becomes half the initial value:

So the energy at that time is

$E(t) = \frac{1}{2} k \left( \frac{A_0}{2} \right)^2 = \frac{1}{2} k \cdot \frac{A_0^2}{4} = \frac{1}{4} \left( \frac{1}{2} k A_0^2 \right) = \frac{1}{4} E_0$ …

When two conducting spheres are connected by a conducting wire, charge will flow between them until they reach electrostatic equilibrium. In electrostatic equilibrium, all points on a conductor, including its surface, are at the same electric potential. Therefore, charge will continue to flow until the electric potential of sphere A becomes equal to the electric potential of sphere B.

Crucially, the total charge in the system remains conserved throughout this process. No charge is lost or gained; it is merely redistributed between the two spheres.

For a conducting sphere of radius $R$ carrying a charge $Q$, the electric potential at its surface (and throughout its volume) is given by:

Let's apply these principles to solve the problem.

1. Identify Initial Conditions and Total Charge:

   We are given the initial charges and radii of the two spheres:

   • Initial charge on sphere A, $Q_A = +90\,\mu\mathrm{C}$
   • Initial charge on sphere B, $Q_B = +90\,\mu\mathrm{C}$
   • Radius of sphere A, $R_A = 8\,\mathrm{cm} = 0.08\,\mathrm{m}$
   • Radius of sphere B, $R_B = 10\,\mathrm{cm} = 0.10\,\mathrm{m}$

   The total initial charge in the system is the sum of the charges on the two spheres:

   $Q_{total} = Q_A + Q_B = 90\,\mu\mathrm{C} + 90\,\mu\mathrm{C} = 180\,\mu\mathrm{C}$.

2. Apply Principle of Potential Equalization:

   When the spheres are connected by a conducting wire, charge flows until their electric potentials become equal. Let the final charges on spheres A and B be $Q_A'$ and $Q_B'$ respectively, and their final potentials be $V_A'$ and $V_B'$.

   At equilibrium, we must have:

   $V_A' = V_B'$

   Using the formula for the potential of a sphere:

   $\frac{kQ_A'}{R_A} = \frac{kQ_B'}{R_B}$

   Since $k$ is a non-zero constant, we can cancel it:

   $\frac{Q_A'}{R_A} = \frac{Q_B'}{R_B}$

3. Apply Principle of Charge Conservation:

   The total charge in the system remains constant. Therefore, the sum of the final charges must equal the total initial charge:

   $Q_A' + Q_B' = Q_{total} = 180\,\mu\mathrm{C}$

4. Solve for Final Charges:

   We now have a system of two equations with two unknowns ($Q_A'$ and $Q_B'$):

   (1) $\frac{Q_A'}{R_A} = \frac{Q_B'}{R_B}$

   (2) $Q_A' + Q_B' = 180\,\mu\mathrm{C}$

   Substitute the given radii into equation (1):

   $\frac{Q_A'}{0.08} = \frac{Q_B'}{0.10}$

   $0.10 Q_A' = 0.08 Q_B'$

   To simplify, multiply by 100:

   $10 Q_A' = 8 Q_B'$

   Divide by 2:

   $5 Q_A' = 4 Q_B'$

   From this, we can express $Q_B'$ in terms of $Q_A'$:

   $Q_B' = \frac{5}{4} Q_A'$

   Now substitute this expression for $Q_B'$ into equation (2): …

Concept and intuition

When two identical conductors with like charges are brought into contact, the total charge redistributes equally between them. The force between them after separation depends on the new charges and the new distance. Here the distance is halved, which quadruples the force from Coulomb’s law, but the charges also change. The given final force (4.5 times the original) lets us solve for the ratio of the original charges.

Step-by-step solution

1. Set up the initial force Let the initial charges be $q_1$ and $q_2$ (both positive, like charges). The initial separation is $r$. Coulomb’s law gives the initial repulsive force:

2. After contact: charge redistribution The balls are identical, so when touched, the total charge $q_1 + q_2$ is shared equally. Each ball gets:

3. After separation: new distance and new force They are moved apart to a distance $r/2$. The new force is:

4. Relate the forces We are told $F' = 4.5 F$. Substitute the expressions:

Cancel $k/r^2$ (nonzero):

5. Solve for the ratio Let $x = q_1/q_2$. Then $q_1 = x q_2$. The equation becomes: $$$$

Why this works

A rectifier (half‑wave or full‑wave) converts AC into a pulsating DC — a voltage that rises and falls with each half‑cycle but never reverses polarity. That’s not yet “steady” enough for most electronics; you need to reduce the ripple. A capacitor placed across the output acts as a temporary reservoir: it charges when the rectified voltage rises, then discharges slowly through the load when the voltage falls, filling in the gaps. This smoothing action yields a much steadier (though not perfectly constant) DC output.

Step‑by‑step reasoning

1. What a rectifier alone does

   A rectifier (using diodes) only flips the negative half‑cycles of AC to positive (or vice versa). The output is a series of humps — still varying from zero to peak. That’s pulsating DC, not steady DC.

2. What a capacitor does in this circuit

   When the rectified voltage rises, the capacitor charges up to the peak voltage. When the rectified voltage falls below the capacitor’s voltage, the capacitor discharges through the load, supplying current and keeping the output voltage from dropping to zero. This “holds up” the voltage between peaks.

3. Result of the smoothing

   The output voltage now has a much smaller variation (ripple) superimposed on a nearly constant DC level. The larger the capacitor (or the lighter the load), the smaller the ripple — the output approaches a steady DC.

4. Eliminating the other options …

Concept & Intuition

For a parallel plate capacitor, the energy stored is $U = \frac{1}{2} C V^2$. Work done to change the voltage is the change in stored energy. Since energy scales as $V^2$, equal voltage increments do not require equal work — the work grows as the voltage increases. The first step (V → 2V) gives a certain work $W$, and the second step (2V → 3V) gives a larger amount. We find the ratio.

Step-by-step reasoning

1. Energy stored at a given voltage The energy stored in a capacitor of capacitance $C$ at voltage $V$ is

2. Work done from V to 2V The work $W$ equals the change in stored energy:

So

3. Work done from 2V to 3V Let this work be $W'$:

So

4. Find $W'$ in terms of $W$ …

Concept and intuition

Surface energy is stored in a liquid surface because molecules at the surface are pulled inward, making the surface behave like a stretched elastic membrane. The energy stored is directly proportional to the surface area:

When a drop breaks into smaller droplets, the total volume stays the same, but the total surface area increases — so extra energy must be supplied. That extra energy is the increase in surface energy we need to compute.

Step-by-step reasoning

1. Find the radius of one small droplet Each small droplet has surface area $A_{\text{small}} = 10^{-7}\ \mathrm{m}^2$. For a sphere, $A = 4\pi r^2$, so

We won’t need $r$ explicitly — we’ll work with areas directly.

2. Total surface area of all 64 droplets

3. Volume conservation: find the original drop’s radius Volume of one small droplet:

Total volume of 64 droplets:

This equals the volume of the original big drop of radius $R$:

4. Surface area of the original drop

But we know $4\pi r^2 = 10^{-7}$, so $64\pi r^2 = 16 \times (4\pi r^2) = 16 \times 10^{-7}\ \mathrm{m}^2$. …

Concept & Intuition

The key idea is charge conservation with sign. When you connect capacitors with opposite polarities, the net charge on the combined system is the algebraic sum of the charges (taking polarity into account). This is not like connecting same-polarity plates, where charges add. Here, positive charge on one plate meets negative charge on the other, so they partially cancel. The final voltage is found by equating the final charge on each capacitor to $Q = CV$, but the total charge is the difference of the initial magnitudes.

Step-by-step reasoning

1. Initial charges on each capacitor Capacitor A: $C$ and voltage $V$ (from the cell).

Capacitor B: $2C$ and voltage $2V$ (from the cell).

2. Polarity matters when connecting The problem says: “plates of opposite polarities touching each other.” That means the positive plate of A is connected to the negative plate of B, and the negative plate of A to the positive plate of B. So the net charge on the entire connected system (both capacitors together) is:

The minus sign tells us the combined system has a net negative charge (excess electrons).

3. After connection, voltage becomes equal In the final steady state, both capacitors have the same voltage $V_f$ across them (they are in parallel). The total capacitance is $C + 2C = 3C$. The final charge on the combination is $Q_{\text{net}} = -3CV$. Using $Q = C_{\text{eq}} V_f$:

The negative sign means the polarity of the combination is opposite to the original polarity of A.

4. Final charges on each capacitor For capacitor A: $Q_A' = C \cdot V_f = C(-V) = -CV$ For capacitor B: $Q_B' = 2C \cdot V_f = 2C(-V) = -2CV$ The magnitude of charge on A is $CV$ (same as initial), but polarity reversed. …

Concept & Intuition

Surface energy is proportional to surface area. When a drop breaks into smaller drops, the total volume stays the same, but the total surface area increases because many small spheres have more combined area than one big sphere. The key is to find how the radius of each small drop relates to the original radius, using volume conservation. Then, compare the total surface area before and after.

Step-by-step reasoning

1. Volume conservation The initial drop has volume $V_i = \frac{4}{3}\pi R^3$. After breaking into $n$ equal drops, each of radius $r$, the total volume is $n \cdot \frac{4}{3}\pi r^3$. Since volume is conserved:

2. Surface area before and after

   Initial surface area (one drop): $A_i = 4\pi R^2$.

   Final total surface area ($n$ drops): $A_f = n \cdot 4\pi r^2 = n \cdot 4\pi \left(\frac{R}{n^{1/3}}\right)^2 = n \cdot 4\pi \frac{R^2}{n^{2/3}} = 4\pi R^2 \cdot n^{1/3}$.

3. Surface energy ratio …

A capacitor is a device designed to store electric charge and electrical energy. It typically consists of two conductors separated by an insulating medium (dielectric). The capacitance ($C$) of a capacitor quantifies its ability to store charge and is defined as the ratio of the magnitude of the charge ($Q$) on either conductor to the magnitude of the potential difference ($V$) between them.

To find the capacitance of any capacitor, the general approach involves these steps:

1. Assume a charge: Place a charge $+Q$ on one conductor and $-Q$ on the other.
2. Calculate the electric field: Determine the electric field $\vec{E}$ in the region between the conductors due to these charges. Gauss's Law is often very useful here, especially for symmetric geometries.
3. Calculate the potential difference: Find the potential difference $V$ between the conductors by integrating the electric field along a path from one conductor to the other. The potential difference is given by $V = -\int \vec{E} \cdot d\vec{l}$.
4. Calculate capacitance: Substitute the expressions for $Q$ and $V$ into the definition $C = Q/V$. The charge $Q$ will cancel out, leaving an expression for $C$ in terms of geometric parameters and fundamental constants.

Let's apply this method to the spherical capacitor.

1. Assume charges on the conductors:

   We have two concentric spherical conductors. Let the inner sphere have radius $R$ and the outer sphere have radius $2R$. We assume a charge $+Q$ is placed on the inner sphere and $-Q$ on the outer sphere.

2. Calculate the electric field between the conductors:

   To find the electric field $\vec{E}$ in the region between the spheres (i.e., for $R < r < 2R$), we use Gauss's Law. We choose a spherical Gaussian surface of radius $r$ concentric with the conductors.

   Due to the spherical symmetry, the electric field $\vec{E}$ will be radial and have the same magnitude at all points on the Gaussian surface.

   Gauss's Law states:

Here, $Q_{enc}$ is the charge enclosed by the Gaussian surface, which is $+Q$ (the charge on the inner sphere). The surface integral simplifies to $E \cdot (4\pi r^2)$ because $\vec{E}$ is perpendicular to the surface and constant in magnitude.

Solving for $E$, we get the electric field magnitude:

This electric field points radially outwards.

3. Calculate the potential difference between the conductors:

The potential difference $V$ between the inner and outer conductors is given by the integral of the electric field from the outer radius to the inner radius. We define $V$ as the potential of the inner sphere relative to the outer sphere, so $V = V_{inner} - V_{outer}$.

$V = -\int_{2R}^{R} \vec{E} \cdot d\vec{l}$ …

When capacitors are connected in series, the key idea is that the charge stored on each capacitor is identical. This is because the same current flows through each during charging, so the same amount of charge accumulates on each plate. The total voltage across the series combination is the sum of the individual voltages, and the equivalent capacitance is given by the reciprocal formula.

Let’s work through the problem step by step.

1. Find the equivalent capacitance. For two capacitors in series:

So

2. Calculate the charge on the equivalent capacitor (which is also the charge on each individual capacitor). The total voltage is $V = 300\,\text{V}$. Using $Q = C_{\text{eq}} V$:

This matches option (A). So (A) is correct.

3. Find the potential difference across each capacitor. For $C_1$:

Actually:

That can’t be right — the total is 300 V. Let’s re-evaluate: $4.8 \times 10^{-4}\,\text{C}$ is $0.00048\,\text{C}$, and $2\,\text{mF} = 0.002\,\text{F}$. So

That seems too small. But wait — the numbers in the problem: $C_1 = 2\,\text{mF}$ means $2 \times 10^{-3}\,\text{F}$, and $Q = 4.8 \times 10^{-4}\,\text{C}$ gives $V_1 = 0.24\,\text{V}$, which is not 60 V. Something is off.

Now with the correct interpretation:

1. Equivalent capacitance:

2. Charge on each capacitor (same in series):

So option (A) is correct.

3. Voltage across $C_1$:

That gives 240 V, not 60 V. But option (B) says 60 V. Let’s check $C_2$:

$V_2 = \frac{Q}{C_2} = \frac{4.8 \times 10^{-4}}{8 \times 10^{-6}} = 60\,\text{V}$ …