To solve the problem step by step, we need to analyze the forces acting on the charged particle and apply the work-energy theorem.
### Step 1: Identify the Given Information
- Charge of the particle, \( q = 4 \, \mu C = 4 \times 10^{-6} \, C \)
- Mass of the particle, \( m = 10 \, mg = 10 \times 10^{-3} \, g = 10 \times 10^{-6} \, kg \)
- Electric field, \( \vec{E} = 4 \hat{i} \, N/C \)
- Magnetic field, \( \vec{B} = 0.2 \hat{k} \, T \)
- Velocity at point \( (x, 3, 0) \), \( \vec{v} = 4 \hat{i} + 3 \hat{j} \)
### Step 2: Calculate the Resultant Velocity
The resultant velocity \( v \) can be calculated using the components:
\[
v = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, m/s
\]
### Step 3: Apply the Work-Energy Theorem
The work done by all forces is equal to the change in kinetic energy:
\[
W_{\text{total}} = \Delta KE = KE_{\text{final}} - KE_{\text{initial}}
\]
Initially, the particle is at rest, so:
\[
KE_{\text{initial}} = 0
\]
The final kinetic energy is:
\[
KE_{\text{final}} = \frac{1}{2} m v^2
\]
Substituting the values:
\[
KE_{\text{final}} = \frac{1}{2} \times (10 \times 10^{-6}) \times (5)^2 = \frac{1}{2} \times 10 \times 10^{-6} \times 25 = 125 \times 10^{-6} \, J = 1.25 \times 10^{-4} \, J
\]
### Step 4: Calculate the Work Done by the Electric Force
The work done by the electric force is given by:
\[
W_E = F_E \cdot d = (qE) \cdot x
\]
Where:
- \( F_E = qE \)
- \( E = 4 \, N/C \)
- \( x \) is the displacement in the \( \hat{i} \) direction.
Calculating the electric force:
\[
F_E = qE = (4 \times 10^{-6}) \times (4) = 16 \times 10^{-6} \, N
\]
Thus, the work done by the electric field is:
\[
W_E = 16 \times 10^{-6} \cdot x
\]
### Step 5: Set Up the Equation
Setting the work done equal to the change in kinetic energy:
\[
16 \times 10^{-6} \cdot x = 1.25 \times 10^{-4}
\]
### Step 6: Solve for \( x \)
Rearranging the equation:
\[
x = \frac{1.25 \times 10^{-4}}{16 \times 10^{-6}} = \frac{1.25}{16} \times 10^{2} = \frac{125}{16} \, m
\]
### Final Answer
Thus, the value of \( x \) is:
\[
x = \frac{125}{16} \, m
\]
