The charge `+q` moving with a velocity `vecV=V_(0)hati` enters a region of uniform electric and magnetic fields with `vecE=4xx10^(3)hatjN//C` and `vecB=0.1kT`. As it moves in the two fields, it deflects along negative y-axis then value of `V_())` must be greater than `x xx10^(4)m//s`. The value of x is:

Explanation

To solve the problem step by step, we need to analyze the forces acting on the charge \( +q \) as it moves through the electric field \( \vec{E} \) and magnetic field \( \vec{B} \).

### Step 1: Identify the forces acting on the charge
The charge \( +q \) experiences two forces:
1. The electric force \( \vec{F_E} \) due to the electric field \( \vec{E} \).
2. The magnetic force \( \vec{F_B} \) due to the magnetic field \( \vec{B} \).

### Step 2: Calculate the electric force
The electric force \( \vec{F_E} \) can be calculated using the formula:
\[ \vec{F_E} = q \vec{E} \]
Given \( \vec{E} = 4 \times 10^3 \hat{j} \, \text{N/C} \), we have:
\[ \vec{F_E} = q (4 \times 10^3 \hat{j}) = 4q \times 10^3 \hat{j} \]

### Step 3: Calculate the magnetic force
The magnetic force \( \vec{F_B} \) is given by:
\[ \vec{F_B} = q (\vec{V} \times \vec{B}) \]
Here, \( \vec{V} = V_0 \hat{i} \) and \( \vec{B} = 0.1 \hat{k} \, \text{T} \). The cross product \( \vec{V} \times \vec{B} \) can be calculated as follows:
\[ \vec{V} \times \vec{B} = (V_0 \hat{i}) \times (0.1 \hat{k}) = V_0 \cdot 0.1 (\hat{i} \times \hat{k}) = V_0 \cdot 0.1 (-\hat{j}) = -0.1 V_0 \hat{j} \]
Thus, the magnetic force becomes:
\[ \vec{F_B} = q (-0.1 V_0 \hat{j}) = -0.1 q V_0 \hat{j} \]

### Step 4: Set up the equation for deflection
For the charge to deflect along the negative y-axis, the net force in the y-direction must be negative. Therefore, we can write:
\[ \vec{F_E} + \vec{F_B} < 0 \]
Substituting the forces we calculated:
\[ 4q \times 10^3 \hat{j} - 0.1 q V_0 \hat{j} < 0 \]

### Step 5: Simplify the inequality
Factoring out \( q \hat{j} \) from the inequality gives:
\[ q (4 \times 10^3 - 0.1 V_0) < 0 \]
Since \( q > 0 \), we can divide by \( q \):
\[ 4 \times 10^3 - 0.1 V_0 < 0 \]
Rearranging this gives:
\[ 0.1 V_0 > 4 \times 10^3 \]
Dividing both sides by \( 0.1 \):
\[ V_0 > \frac{4 \times 10^3}{0.1} = 4 \times 10^4 \, \text{m/s} \]

### Step 6: Determine the value of \( x \)
The problem states that \( V_0 \) must be greater than \( x \times 10^4 \, \text{m/s} \). From our calculation, we have:
\[ V_0 > 4 \times 10^4 \, \text{m/s} \]
Thus, we can conclude that:
\[ x = 4 \]

### Final Answer
The value of \( x \) is \( 4 \).