The oscillating magnetic field in a plane electromagnetic wave is given by $$B_y = 5 \times 10^{-6} \sin[1000\pi(5x - 4 \times 10^8 t)] \text{ T}$$. The amplitude of electric field will be

The oscillating magnetic field in a plane electromagnetic wave is given by: $$B_y = 5 \times 10^{-6} \sin[1000\pi(5x - 4 \times 10^8 t)] \text{ T}$$, and we need to find the amplitude of the electric field.

Since the amplitude of the magnetic field is the coefficient of the sine function, we have $$B_0 = 5 \times 10^{-6} \text{ T}$$.

From the general form of a travelling wave, $$B = B_0 \sin(kx - \omega t)$$, expanding the given argument yields $$1000\pi(5x - 4 \times 10^8 t) = 5000\pi \cdot x - 4\pi \times 10^{11} \cdot t$$. Therefore, $$k = 5000\pi \text{ m}^{-1}$$ and $$\omega = 4\pi \times 10^{11} \text{ rad s}^{-1}$$.

Since the wave speed is given by $$c = \dfrac{\omega}{k}$$, substituting the values of $$\omega$$ and $$k$$ gives $$c = \dfrac{4\pi \times 10^{11}}{5000\pi} = \dfrac{4 \times 10^{11}}{5 \times 10^3} = 8 \times 10^7 \text{ m s}^{-1}$$.

For an electromagnetic wave, the amplitudes of the electric and magnetic fields are related by $$E_0 = c \times B_0$$, where $$c$$ is the wave speed in the medium. Hence, $$E_0 = 8 \times 10^7 \times 5 \times 10^{-6} = 40 \times 10^{7-6} = 40 \times 10^1 = 400 \text{ V m}^{-1}$$, which is $$4 \times 10^2 \text{ V m}^{-1}$$.

Note that the wave speed here is $$8 \times 10^7 \text{ m s}^{-1}$$, which is less than the speed of light in vacuum ($$3 \times 10^8 \text{ m s}^{-1}$$). This indicates the wave is travelling in a medium, and the relation $$E_0 = cB_0$$ still holds where $$c$$ is the wave speed in that medium.

Therefore, the correct answer is Option D: $$4 \times 10^2 \text{ V m}^{-1}$$.