To solve this problem, we need to analyze the motion of a charged particle in the presence of both electric and magnetic fields. The particle starts from rest, and we want to determine its speed when it reaches the point (3, 2, 1). Let's break this down step by step.

Understanding the Forces Acting on the Charge

The particle with charge \( q \) and mass \( m \) experiences two forces due to the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{B} \).

• The electric force \( \mathbf{F}_E \) is given by:

  F_E = q \mathbf{E}

• The magnetic force \( \mathbf{F}_B \) is given by:

  F_B = q (\mathbf{v} \times \mathbf{B})

Initially, the particle is at rest, so its velocity \( \mathbf{v} = 0 \). This means that at the start, the magnetic force does not contribute to the motion. The only force acting on the particle is the electric force.

Calculating the Electric Force

Given the electric field \( \mathbf{E} = E_0 \mathbf{i} + 2E_0 \mathbf{j} \), we can calculate the electric force:

F_E = q (E_0 \mathbf{i} + 2E_0 \mathbf{j}) = qE_0 \mathbf{i} + 2qE_0 \mathbf{j}

Finding the Acceleration

Using Newton's second law, we can find the acceleration \( \mathbf{a} \) of the particle:

\(\mathbf{F} = m \mathbf{a} \Rightarrow \mathbf{a} = \frac{\mathbf{F}}{m}\)

Substituting the electric force:

\(\mathbf{a} = \frac{qE_0}{m} \mathbf{i} + \frac{2qE_0}{m} \mathbf{j}\)

Using Kinematics to Find the Speed

Now, we need to determine the speed of the particle when it reaches the point (3, 2, 1). The particle moves under constant acceleration, so we can use the kinematic equation:

v^2 = u^2 + 2a s

Since the particle starts from rest, \( u = 0 \), and the equation simplifies to:

v^2 = 2a s

Calculating the Displacement

The displacement vector from the origin to the point (3, 2, 1) is:

\( \mathbf{s} = 3 \mathbf{i} + 2 \mathbf{j} + 1 \mathbf{k} \)

The magnitude of the displacement \( s \) is:

\( s = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14} \)

Finding the Total Acceleration

The total acceleration \( a \) can be calculated using the components:

\( a = \sqrt{\left(\frac{qE_0}{m}\right)^2 + \left(\frac{2qE_0}{m}\right)^2} = \frac{qE_0}{m} \sqrt{1 + 4} = \frac{qE_0}{m} \sqrt{5} \)

Putting It All Together

Now we can substitute \( a \) and \( s \) back into the kinematic equation:

v^2 = 2 \left(\frac{qE_0}{m} \sqrt{5}\right) \sqrt{14}

Thus, we find:

v^2 = \frac{2qE_0 \sqrt{70}}{m}

Taking the square root gives:

v = \sqrt{\frac{2qE_0 \sqrt{70}}{m}} = \frac{qE_0}{m} \sqrt{14} \sqrt{5} = \frac{qE_0}{m} \cdot \sqrt{70}

Final Result

Now, we can compare this with the options provided. The speed of the particle when it reaches the point (3, 2, 1) does not match any of the given options directly. Therefore, the answer is:

d) none of these