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Step-by-Step Solution
Step 1: Write down the given electric field
The electric field of the plane electromagnetic wave is
$ \overrightarrow{E} = E_{0}\,\frac{\widehat{i} + \widehat{j}}{\sqrt{2}} \cos(kz + \omega t). $
Step 2: Evaluate the electric field at the specified location and time
At time $ t = 0 $ and at position
$ \bigl(x,y,z\bigr) = \Bigl(0,0,\frac{\pi}{k}\Bigr), $
we substitute into the argument of the cosine:
$ kz + \omega t = k\Bigl(\frac{\pi}{k}\Bigr) + \omega (0) = \pi. $
Since $ \cos(\pi) = -1, $ the electric field becomes
$ \overrightarrow{E} \bigl(t=0,\,z=\tfrac{\pi}{k}\bigr)
= E_{0}\,\frac{\widehat{i} + \widehat{j}}{\sqrt{2}} \bigl[-1\bigr]
= -E_{0}\,\frac{\widehat{i} + \widehat{j}}{\sqrt{2}}.
$
Step 3: Express the net force on the charged particle
The force on a charged particle (charge $q$) in an electromagnetic wave is given by
$ \overrightarrow{F} = q\Bigl(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B}\Bigr). $
We are told the instantaneous velocity of the particle at that moment is
$ \overrightarrow{v} = v_{0}\,\widehat{k}. $
Step 4: Consider the magnetic force term
In an electromagnetic wave traveling along some direction (say the $z$-axis),
$ \overrightarrow{E} $ and $ \overrightarrow{B} $ are perpendicular to each other and also perpendicular to the direction of propagation. Because the velocity $ \overrightarrow{v} $ is along $ \widehat{k} $, the magnetic force term
$ q\bigl(\overrightarrow{v} \times \overrightarrow{B}\bigr) $
lies in a direction perpendicular to $ \overrightarrow{v} $ and $ \overrightarrow{B}$.
However, one can show (or from the known relations of electromagnetic waves) that at this instant, the net effect of $ \overrightarrow{v} \times \overrightarrow{B} $ does not cancel or override the direction of $ \overrightarrow{E} $; effectively, the force due to $ \overrightarrow{v} \times \overrightarrow{B} $ aligns with $ \pm \overrightarrow{E} $. In this situation, the dominant direction of the net force is dictated by $ \overrightarrow{E} $.
Step 5: Determine the final direction of the net force
Because $ \overrightarrow{E} $ at the given point and time is
$ -E_{0}\,\frac{\widehat{i} + \widehat{j}}{\sqrt{2}}, $
the force on a positively charged particle is antiparallel to
$ \frac{\widehat{i} + \widehat{j}}{\sqrt{2}}. $
Hence, the correct answer is antiparallel to
$ \frac{\widehat{i} + \widehat{j}}{\sqrt{2}}. $
