An electrical charge `2xx10^-8` C is placed at the point `(1,2,4)` m. At the point `(4,2,0) `m, the electric

Options

A. potential will be 36 V
B. field will along y-axis
C. field will increase if the space between the points is filled with a dielectric
D. All of the above

Explanation

To solve the problem, we need to find the electric potential and electric field at the point (4, 2, 0) m due to a point charge of \(2 \times 10^{-8}\) C located at (1, 2, 4) m.

### Step 1: Identify the position vectors
Let \( r_1 = (1, 2, 4) \) m be the position of the charge and \( r_2 = (4, 2, 0) \) m be the point where we want to find the electric potential and electric field.

### Step 2: Calculate the position vector \( \mathbf{r} \)
The position vector \( \mathbf{r} \) from the charge to the point is given by:
\[ \mathbf{r} = \mathbf{r_2} - \mathbf{r_1} = (4 - 1) \hat{i} + (2 - 2) \hat{j} + (0 - 4) \hat{k} = 3 \hat{i} + 0 \hat{j} - 4 \hat{k} \]
Thus, \( \mathbf{r} = 3 \hat{i} - 4 \hat{k} \).

### Step 3: Calculate the magnitude of \( \mathbf{r} \)
The magnitude of \( \mathbf{r} \) is given by:
\[ |\mathbf{r}| = \sqrt{(3)^2 + (0)^2 + (-4)^2} = \sqrt{9 + 0 + 16} = \sqrt{25} = 5 \text{ m} \]

### Step 4: Calculate the unit vector \( \hat{r} \)
The unit vector \( \hat{r} \) in the direction of \( \mathbf{r} \) is given by:
\[ \hat{r} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{3 \hat{i} - 4 \hat{k}}{5} = \frac{3}{5} \hat{i} + 0 \hat{j} - \frac{4}{5} \hat{k} \]

### Step 5: Calculate the electric potential \( V \)
The electric potential \( V \) due to a point charge is given by:
\[ V = k \frac{Q}{|\mathbf{r}|} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( Q = 2 \times 10^{-8} \, \text{C} \).
Substituting the values:
\[ V = 9 \times 10^9 \frac{2 \times 10^{-8}}{5} = 9 \times 10^9 \times \frac{2}{5} \times 10^{-8} = 36 \, \text{V} \]

### Step 6: Calculate the electric field \( \mathbf{E} \)
The electric field \( \mathbf{E} \) due to a point charge is given by:
\[ \mathbf{E} = k \frac{Q}{|\mathbf{r}|^2} \hat{r} \]
Substituting the values:
\[ \mathbf{E} = 9 \times 10^9 \frac{2 \times 10^{-8}}{5^2} \hat{r} = 9 \times 10^9 \frac{2 \times 10^{-8}}{25} \hat{r} \]
Calculating \( \frac{2 \times 10^{-8}}{25} = 8 \times 10^{-10} \):
\[ \mathbf{E} = 9 \times 10^9 \times 8 \times 10^{-10} \hat{r} = 7.2 \hat{r} \, \text{N/C} \]
Now substituting \( \hat{r} = \frac{3}{5} \hat{i} - \frac{4}{5} \hat{k} \):
\[ \mathbf{E} = 7.2 \left( \frac{3}{5} \hat{i} - \frac{4}{5} \hat{k} \right) = 4.32 \hat{i} - 5.76 \hat{k} \, \text{N/C} \]

### Final Answers
- Electric Potential \( V = 36 \, \text{V} \)
- Electric Field \( \mathbf{E} = 4.32 \hat{i} - 5.76 \hat{k} \, \text{N/C} \)