Electric field in space is given by `vec(E(t)) = E_0 (i+j)/sqrt2 cos(omegat+Kz)`. A positively charged particle at `(0, 0, pi/K)` is given velocity `v_0 hatk` at t = 0. Direction of force acting on particle is

Options

A. Option1 F = 0
B. Option2 antiparallel to `(hati+hatj)/sqrt2`
C. Option3 parallel to `(hati+hatj)/sqrt2`
D. Option4 `hatk`

Explanation

To solve the problem, we need to determine the direction of the force acting on a positively charged particle in a given electric field. The electric field is defined as:

\[ \vec{E}(t) = E_0 \frac{(i + j)}{\sqrt{2}} \cos(\omega t + Kz) \]

The particle is located at the point \((0, 0, \frac{\pi}{K})\) and has an initial velocity of \(\vec{v} = v_0 \hat{k}\) at \(t = 0\).

### Step 1: Evaluate the Electric Field at \(t = 0\) and \(z = \frac{\pi}{K}\)

Substituting \(t = 0\) and \(z = \frac{\pi}{K}\) into the electric field equation:

\[ \vec{E}(0) = E_0 \frac{(i + j)}{\sqrt{2}} \cos(0 + K \cdot \frac{\pi}{K}) = E_0 \frac{(i + j)}{\sqrt{2}} \cos(\pi) \]

Since \(\cos(\pi) = -1\):

\[ \vec{E}(0) = -E_0 \frac{(i + j)}{\sqrt{2}} \]

### Step 2: Calculate the Electric Force on the Particle

The force due to the electric field on the charged particle is given by:

\[ \vec{F}_E = q \vec{E} \]

Substituting \(\vec{E}(0)\):

\[ \vec{F}_E = q \left(-E_0 \frac{(i + j)}{\sqrt{2}}\right) = -\frac{q E_0}{\sqrt{2}} (i + j) \]

### Step 3: Determine the Magnetic Field Direction

Since the electric field is time-varying, a magnetic field \(\vec{B}\) is generated. The direction of the magnetic field can be determined using the right-hand rule. The electric field \(\vec{E}\) and the direction of wave propagation (which is along \(\hat{k}\)) are perpendicular to each other.

Given that \(\vec{E}\) is in the \(-i\) and \(-j\) directions, the magnetic field \(\vec{B}\) will be in the \(\hat{k}\) direction (out of the page).

### Step 4: Calculate the Magnetic Force

The magnetic force on the charged particle is given by:

\[ \vec{F}_B = q \vec{v} \times \vec{B} \]

Substituting the values:

\[ \vec{F}_B = q (v_0 \hat{k}) \times \vec{B} \]

Assuming \(\vec{B}\) is in the \(\hat{i}\) direction (perpendicular to both \(\hat{E}\) and \(\hat{k}\)), we can use the right-hand rule to find the direction of \(\vec{F}_B\).

### Step 5: Combine the Forces

The total force acting on the particle is:

\[ \vec{F}_{\text{total}} = \vec{F}_E + \vec{F}_B \]

Since both forces will have components in the \(-i\) and \(-j\) directions, the net force will also point in the direction of \(-\frac{(i + j)}{\sqrt{2}}\).

### Conclusion

The direction of the force acting on the particle is in the direction of the electric field at \(t = 0\), which is:

\[ -\frac{(i + j)}{\sqrt{2}} \]