Exams › JEE Main › Physics

A parallel-plate air capacitor, each plate of area S = 100 cm², is connected in series in an ac circuit. The sinusoidal current amplitude in the leads is Iₘ = 1.0 mA at angular frequency omega = 1.6 x 10⁷ rad/s. Find the amplitude of the electric field strength between the plates.

1. 57 V/cm
2. 7.2 V/cm
3. 28 V/cm
4. 115 V/cm

Correct answer: 7.2 V/cm

Solution

Equating lead current to displacement current: Iₘ = e0 S omega Eₘ, so Eₘ = Iₘ/(e0 S omega) = 1.0x10⁻³ /(8.85x10⁻¹² * 1.0x10⁻² * 1.6x10⁷) ~ 706 V/m ~ 7.1 V/cm. This matches Irodov's result of about 7 V/cm.

Related JEE Main Physics questions

• A plane electromagnetic wave falls on a material surface. If it transfers momentum p and energy E, which statement is true?
• For an electromagnetic wave moving in vacuum, the electric field is expressed as E = E₀ sin(kx − ωt). Which of the following quantities does not depend on the wavelength?
• When an electromagnetic wave enters the ionised region of the ionosphere, the moving electron cloud gives rise to a space current, while the electric field also produces a capacitive displacement current. In this situation, what is the phase relation between the space current and the displacement current?
• The rms electric field associated with sunlight is 720 N/C. What is the mean total energy density of this electromagnetic wave?
• The electric field component of an electromagnetic wave in a medium is given by Eₓ = 0, E_y = 2.5 N/C cos[(2π × 10⁶ rad/m)x - (π × 10⁻² rad/s)t] ŷ, and E_z = 0. The wave is:
• A source of monochromatic green light radiates equally in every direction. If it converts only 3% of the 100 W electrical power it uses into electromagnetic radiation, then the electric-field amplitude of the emitted wave at a point 5 m away from the source is closest to:

⚔️ Practice JEE Main Physics free + battle 1v1 →