Question

An electric field of 106 N/C acts on an electron, resulting in an acceleration of 1.3 ✕ 1017 m/s2 for a short time. What is the magnitude of the radiative electric field observed at a location a distance of 4 cm away along a line perpendicular to the direction of the acceleration?

Answer 1

Answer:

5.20535×10⁻⁸ V/m

Explanation:

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a = Acceleration of electron = 1.3×10¹⁷ m/s²

r = Radiative distance = 4 cm = 0.04 m

c = Speed of light

q = Charge of electron = 1.6022×10⁻¹⁹ C

Radiative electric field

$E_{radiative}=-\frac{1}{4\pi \epsilon_{0}}\frac{qa}{c^2r}\\\Rightarrow E_{radiative}=-\frac{1}{4\pi 8.845\times 10^{-12}}\frac{1.6022\times 10^{-19}\times 1.3\times 10^{17}}{(3\times 10^8)^2\times 0.04}=-5.20535\times 10^{-8}$

∴ Magnitude of the radiative electric field observed is 5.20535×10⁻⁸ V/m