A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by $$\vec{B} = 5 \times 10^{-8}\hat{j}$$ T. The corresponding electric field $$\vec{E}$$ is (speed of light = $$3 \times 10^8$$ m s$$^{-1}$$)

We have a plane electromagnetic wave propagating in vacuum along the $$+z$$-direction. In such a wave the electric field $$\vec{E}$$, the magnetic field $$\vec{B}$$ and the direction of propagation are all mutually perpendicular.

For any electromagnetic wave in vacuum the magnitudes of the fields are related by the formula

$$|\vec{E}| = c\,|\vec{B}|,$$

where $$c = 3 \times 10^{8}\ \text{m s}^{-1}$$ is the speed of light.

The magnetic field at the specified instant is

$$\vec{B} = 5 \times 10^{-8}\,\hat{j}\ \text{T}.$$

Substituting this value into the magnitude relation gives

$$|\vec{E}| = (3 \times 10^{8})\,(5 \times 10^{-8}) = 15\ \text{V m}^{-1}.$$

Next we determine the direction of $$\vec{E}$$. The Poynting vector $$\vec{S}$$, which is proportional to $$\vec{E} \times \vec{B},$$ must point along the $$+z$$-axis. Using the right-hand rule:

$$\hat{i} \times \hat{j} = \hat{k}.$$

Therefore, if $$\vec{E}$$ is along $$+\hat{i}$$ and $$\vec{B}$$ is along $$+\hat{j},$$ their cross product points along $$+\hat{k},$$ matching the given propagation direction. Choosing $$-\hat{i}$$ for $$\vec{E}$$ would give a cross product along $$-\hat{k},$$ which is not acceptable here.

Hence the electric field vector is

$$\vec{E} = 15\,\hat{i}\ \text{V m}^{-1}.$$

Hence, the correct answer is Option D.