An electric charge `2 xx 10^(-8) C` is placed at the point `(1,2,4)`. At the point `(3,2,1)` , the electric
(i) potential will be `50 V`
(ii) field will have no `Y`-component
(iii) field will increase by a factor `K` if the space between the points is filled with a dielectric of constant `K`
(iv) field will be along `Y`-axis

Options

A. field will increase by a factor K if the space between the points is filled with a dielectric of dielectric constant K
B. field will be along the y-axis.
C. potential will be 49.9 V.
D. field will have no y-components.

Explanation

To solve the problem step by step, we will analyze the given information and calculate the necessary values.

### Step 1: Identify the Charge and Positions
We have a charge \( Q = 2 \times 10^{-8} \, \text{C} \) located at point \( A(1, 2, 4) \) and we need to evaluate the electric potential and electric field at point \( B(3, 2, 1) \).

### Step 2: Calculate the Position Vector
The position vector from point \( A \) to point \( B \) can be calculated as:
\[ \vec{r} = \vec{r_B} - \vec{r_A} = (3 \hat{i} + 2 \hat{j} + 1 \hat{k}) - (1 \hat{i} + 2 \hat{j} + 4 \hat{k}) \]
\[ \vec{r} = (3 - 1) \hat{i} + (2 - 2) \hat{j} + (1 - 4) \hat{k} = 2 \hat{i} + 0 \hat{j} - 3 \hat{k} \]

### Step 3: Calculate the Magnitude of the Distance
The magnitude of the distance \( R \) between points \( A \) and \( B \) is given by:
\[ R = |\vec{r}| = \sqrt{(2)^2 + (0)^2 + (-3)^2} = \sqrt{4 + 0 + 9} = \sqrt{13} \, \text{m} \]

### Step 4: Calculate the Electric Potential at Point B
The electric potential \( V \) at point \( B \) due to the charge \( Q \) is given by:
\[ V_B = \frac{kQ}{R} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

Substituting the values:
\[ V_B = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{\sqrt{13}} = \frac{180 \times 10^1}{\sqrt{13}} \approx 49.92 \, \text{V} \]
Thus, we can round this to \( 50 \, \text{V} \).

### Step 5: Analyze the Electric Field
The electric field \( \vec{E} \) due to a point charge is given by:
\[ \vec{E} = \frac{kQ}{R^3} \vec{r} \]
Substituting \( \vec{r} = 2 \hat{i} + 0 \hat{j} - 3 \hat{k} \):
\[ \vec{E} = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{(\sqrt{13})^3} (2 \hat{i} + 0 \hat{j} - 3 \hat{k}) \]
Since there is no \( \hat{j} \) component in \( \vec{r} \), the \( y \)-component of the electric field will be zero.

### Step 6: Effect of Dielectric
When a dielectric material with constant \( K \) is introduced, the electric field \( E \) decreases by a factor of \( K \):
\[ E' = \frac{E}{K} \]
Thus, if the space between the points is filled with a dielectric, the electric field will decrease, not increase.

### Conclusion
Based on the calculations:
1. The electric potential at point \( B \) is \( 50 \, \text{V} \) (Correct).
2. The electric field has no \( y \)-component (Correct).
3. The electric field will decrease by a factor \( K \) if a dielectric is introduced (Incorrect).
4. The electric field will not be along the \( y \)-axis (Incorrect).

### Final Answers
- (i) True
- (ii) True
- (iii) False
- (iv) False