Q.Figure 8.5 shows a capacitor made of two circular plates each of radius $12\ \text{cm}$, and separated by $5.0\ \text{cm}$. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15\ \text{A}$.

Concept

Displacement current, introduced by Maxwell to complete Ampère's law, has density $J_d=\varepsilon_0\dfrac{\partial E}{\partial t}$. It is not a flow of charge: it arises purely from a changing electric field and, like a real current, produces a magnetic field. Because it carries no charge carriers and no resistance, it has no $J=\sigma E$ (Ohm's-law) relation.

Solution

Test each statement:

1. A — "exists only when $E$ changes with time": true, since $J_d\propto \partial E/\partial t$; a static field gives zero displacement current.
2. B — "obeys Ohm's law": false. Ohm's law $J=\sigma E$ applies to conduction currents; displacement current depends on $\partial E/\partial t$, not on $E$ through a conductivity.
3. C — "produces a magnetic field just like conduction current": true — this is precisely Maxwell's correction, $\nabla\times B=\mu_0\big(J_c+\varepsilon_0\partial E/\partial t\big)$.
4. D — "unlike conduction current, no electron flow through a conductor": true — it exists even in vacuum.

The only incorrect statement is B. Among the answer choices, the one that selects statement B is option (A).

Step 1 — The concept: energy stored in a capacitor

At any instant when the potential difference across a capacitor is $V$,

Note this is the instantaneous energy — it depends on the capacitor's present voltage $V$, not on the battery's emf. The emf of $10\ \text{V}$ only tells us the capacitor is still charging (it has reached $4\ \text{V}$ of an eventual $10\ \text{V}$).

Step 2 — Differentiate with respect to time

$C$ is a constant of the capacitor, and $V$ is the only time-dependent quantity, so by the chain rule:

Step 3 — Substitute the instantaneous values

Step 4 — Check the units

$\text{F}\times\text{V}\times\text{V s}^{-1} = \text{(C/V)}\cdot\text{V}\cdot\text{V s}^{-1} = \text{C V s}^{-1} = \text{J s}^{-1} = \text{W}$. ✓

The answer is not zero (option C): the capacitor is still charging, since $V = 4\ \text{V} < 10\ \text{V}$ and $dV/dt \neq 0$.

ANSWER: A

The key idea is that displacement current is not a separate physical current; it is the term Maxwell added to make the total current continuous in circuits containing capacitors. In a parallel plate capacitor, the displacement current between the plates exactly equals the conduction current in the connecting wires at every moment. This is because the changing electric field between the plates produces a displacement current density $\mathbf{J}_d = \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$, and integrating over the plate area gives the same current as flows into the capacitor.

Thus, if the rms conduction current in the wire is $4\ \mu\text{A}$, the rms displacement current between the plates is also $4\ \mu\text{A}$. No further calculation is needed — the problem tests this conceptual equality.

Let’s verify with the given numbers to be thorough:

1. Given data

   • Capacitance: $C = 400\ \text{pF} = 400 \times 10^{-12}\ \text{F}$
   • AC source voltage (rms): $V_{\text{rms}} = 100\ \text{V}$
   • Angular frequency: $\omega = 100\ \text{rad/s}$
   • RMS current: $I_{\text{rms}} = 4\ \mu\text{A} = 4 \times 10^{-6}\ \text{A}$

2. Check consistency

   For a capacitor, the impedance is $X_C = \frac{1}{\omega C}$.

The rms current from Ohm’s law for AC would be:

This matches the given current, confirming the circuit is purely capacitive and the displacement current equals the conduction current.

3. Conclusion The displacement current’s rms value is exactly the same as the conduction current’s rms value: $4\ \mu\text{A}$.

The key idea is simple: electromagnetic waves are produced by accelerating charges. A charge at rest or moving with constant velocity (in a straight line) produces only static or steady fields — no radiation. But when a charge accelerates, its electric and magnetic fields change in time, and those changes propagate outward as electromagnetic waves.

Let’s examine each option carefully.

1. Option (A): Moving with a constant velocity

   A charge moving at constant velocity has zero acceleration. Its electric field is steady (though it moves with the charge), and its magnetic field is also steady. No time-varying fields are produced, so no electromagnetic waves are emitted.

   Watch out

   A common mistake is to think that any moving charge radiates. Only accelerating charges radiate — constant velocity is not enough.

2. Option (B): Moving in a circular orbit

   Circular motion, even at constant speed, involves continuous acceleration toward the centre (centripetal acceleration). This acceleration causes the charge’s velocity direction to change constantly, producing time-varying electric and magnetic fields. These fields propagate away as electromagnetic waves. This is exactly how synchrotron radiation is produced — electrons in a circular accelerator emit intense EM waves.

3. Option (C): At rest

   A stationary charge produces only a static electric field. No magnetic field at all (unless there is a current, but here the charge is alone and at rest). No time variation means no wave.

4. Option (D): Moving parallel to the magnetic field

   A charge moving parallel to a uniform magnetic field experiences no magnetic force ($\vec{F} = q\vec{v} \times \vec{B} = 0$ when $\vec{v} \parallel \vec{B}$). So it continues with constant velocity — no acceleration, no radiation.

   Tip

   Even if the charge moves in a magnetic field, it’s the acceleration (due to a perpendicular component of velocity) that causes radiation, not the mere presence of the field.

The key idea is that in an AC circuit, power is not simply $V^2/R$ because the inductor stores and returns energy without dissipation. Only the resistor consumes real power, and the phase difference between voltage and current reduces the net power.

For a series L-R circuit driven by an AC source of rms voltage $V$ and angular frequency $\omega$, the impedance is $Z = \sqrt{R^2 + (\omega L)^2}$. The rms current is $I = V/Z$. The power factor is $\cos\phi = R/Z$, where $\phi$ is the phase angle by which current lags voltage. Average power dissipated is $P = V I \cos\phi$.

1. Write the impedance:

2. The rms current:

3. The power factor:

4. Average power:

Step 1 — Why the formula is just $P = VI$.

In a general AC circuit the average power is

where $\cos\phi$ is the power factor. A bulb's filament is a pure resistance — no inductance or capacitance — so voltage and current are in phase, $\phi = 0$ and $\cos\phi = 1$. The $50~\mathrm{Hz}$ therefore plays no role at all; it is given only as a distractor.

Step 2 — Substitute the ratings.

Ratings on a bulb are always rms (average) values.

Step 3 — Simplify the fraction.

Cross-check. Resistance $R = V^2/P = 220^2/100 = 484\ \Omega$, and $I = V/R = 220/484 = 5/11$ A. ✓

ANSWER: A