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 $m$.

A capacitor is made of two circular plates of radius R each, separated by a distance d ll R. The capacitor is

NCERT Exemplar · Q27

R

Uttarakhand UbseLong· 5mImportance★★★★★

86% · 48/56 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Capacitor Energy Storage

Capacitor Energy Storage

The Intuition

Charging a capacitor is like piling sand onto a growing heap. The first grain of charge lands on an empty plate easily. But every later bit of positive charge must be pushed onto a plate that is already positive, and it resists. So more and more work is needed as the plate fills up. All of that work does not disappear — it is stored in the capacitor as electrostatic potential energy, ready to be released later.

Building the Formula

Suppose at some instant during charging the capacitor already holds charge $q$ , so the voltage across it at that moment is $v = q/C$ . Moving one more small charge $dq$ onto the plate costs work:

dW = v\,dq = \frac{q}{C}\,dq

Adding up (integrating) all these small contributions as the charge builds from $0$ to a final value $Q$ gives the total work done:

W = \int_0^{Q} \frac{q}{C}\,dq = \frac{1}{C}\cdot\frac{Q^2}{2} = \frac{Q^2}{2C}

This work is exactly the energy $U$ stored in the charged capacitor.

Three Equivalent Forms

Using $Q = CV$ , the same stored energy can be written three ways — pick whichever matches the quantities you know:

U = \frac{Q^2}{2C} = \frac{1}{2}QV = \frac{1}{2}CV^2

Use $\dfrac{Q^2}{2C}$ when the charge is fixed (capacitor disconnected from the source).
Use $\dfrac{1}{2}CV^2$ when the voltage is fixed (capacitor stays connected to a battery).

Watch out

The factor of $\tfrac{1}{2}$ is essential. A common error is writing $U = QV$ . That would only be true if the full voltage $V$ acted while all the charge moved — but the voltage climbs steadily from $0$ to $V$ as the plates fill, so the effective average voltage is $V/2$ , giving $U = \tfrac{1}{2}QV$ .

Where the Energy Lives — Energy Density

The energy is stored in the electric field occupying the space between the plates, not on the plates themselves. For a parallel-plate capacitor this leads to a general result: energy stored per unit volume of field is

u = \frac{1}{2}\,\varepsilon_0 E^2

where $E$ is the field strength. Wherever an electric field exists, energy is stored there, with density proportional to $E^2$ .

A Quick Example

A $10\ \mu\text{F}$ capacitor is charged to $100\ \text{V}$ . The stored energy is:

U = \frac{1}{2}CV^2 = \frac{1}{2}\times(10\times10^{-6})\times(100)^2 = 0.05\ \text{J}

That 0.05 J can be released almost instantly — which is exactly how a camera flash works: charge slowly, discharge fast. …

Start your 14-day free trial to unlock the full solution →

Because the disc is a conductor sitting on the plate, the electrostatic force lifting it comes from the surface-charge pressure $\sigma^2/(2\varepsilon_0)$ acting on its own induced charge — not the naive $\sigma E$ — giving $V_{\min}=\dfrac{d}{r}\sqrt{\dfrac{2mg}{\pi\varepsilon_0}}$ .

Setting up the field and the induced charge

Since $d\ll R$ , the field between the plates is uniform:

E=\frac{V}{d}.

The thin conducting disc ( $t\ll r$ ) sits flush on the bottom plate, so it is at the same potential as that plate and effectively becomes part of the conducting boundary. Just like the rest of the bottom plate, its exposed top face carries an induced surface charge density

\sigma=\varepsilon_0 E=\frac{\varepsilon_0 V}{d},

found from the standard boundary condition that the field just outside a conductor's surface is $E=\sigma/\varepsilon_0$ .

The subtle point: the disc cannot pull on itself

Here is where the naive approach goes wrong. It is tempting to say "force = charge × field = $q\times E$ ", using the full field $E$ between the plates. But a charge element sitting on the disc's own surface cannot feel a force from its own field — a charge cannot exert a net force on itself. The force it actually feels comes only from the field due to everything else (the rest of the disc's charge plus the top plate).

Right at the conductor's surface, the total field jumps from $0$ (just inside the conductor) to $E=\sigma/\varepsilon_0$ (just outside). The field "due to everything else" (excluding this element's own contribution) at that location is the average of these two values:

E_{\text{other}}=\frac{0+E}{2}=\frac{E}{2}.

This is the well-known result that a charged conductor's surface experiences an outward electrostatic pressure

P=\sigma\cdot\frac{E}{2}=\frac{\sigma^2}{2\varepsilon_0}

per unit area — half of what the naive $\sigma E$ would give. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ26 — A parallel plate capacitor is filled by a dielectric whose relative pe…→Next questionQ28 — (a) In a quark model of elementary particles, a neutron is made of one…

Concept understanding — Capacitor Energy Storage

Capacitor Energy Storage

The Intuition

Building the Formula

Three Equivalent Forms

Where the Energy Lives — Energy Density

A Quick Example

Setting up the field and the induced charge

The subtle point: the disc cannot pull on itself

Method: Force on a Conductor Due to Its Own Induced Surface Charge

Steps

Setting up the field and the induced charge

The subtle point: the disc cannot pull on itself

🎓Unlock everything free for 14 days

Unlock everything free for 14 days