A point charge `Q` is placed at origin. Let `vecE_(A), vec E_(B), and vecE_(C)` be the electirc field at three points A (1,2,3), B (1,1,1), and C(2,2,2) due to charge `Q`. Then

Options

A. `vecE_(A)_|_vecE_(B)`
B. `vecE_(A)||vecE_(B)`
C. `|vecE_(B)| = 4|vecE_(C)|`
D. `vecE_(B) = 16 | vecE_(C)|`

Explanation

To solve the problem of finding the relationship between the electric fields at points A, B, and C due to a point charge Q placed at the origin, we will follow these steps:

### Step 1: Define the position vectors
The position vectors of the points A, B, and C with respect to the origin (where the charge Q is located) are:
- Point A: \(\vec{r}_A = \hat{i} + 2\hat{j} + 3\hat{k}\)
- Point B: \(\vec{r}_B = \hat{i} + \hat{j} + \hat{k}\)
- Point C: \(\vec{r}_C = 2\hat{i} + 2\hat{j} + 2\hat{k}\)

### Step 2: Calculate the magnitudes of the position vectors
Next, we calculate the magnitudes of these position vectors:
- Magnitude of \(\vec{r}_A\):
\[ R_A = |\vec{r}_A| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

- Magnitude of \(\vec{r}_B\):
\[ R_B = |\vec{r}_B| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

- Magnitude of \(\vec{r}_C\):
\[ R_C = |\vec{r}_C| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} \]

### Step 3: Calculate the electric fields at points A, B, and C
The electric field \(\vec{E}\) due to a point charge \(Q\) at a distance \(R\) is given by:
\[ \vec{E} = \frac{kQ}{R^2} \hat{r} \]
where \(\hat{r}\) is the unit vector in the direction of the position vector.

- Electric field at point A:
\[ \vec{E}_A = \frac{kQ}{R_A^2} \hat{r}_A = \frac{kQ}{14} \frac{\vec{r}_A}{|\vec{r}_A|} = \frac{kQ}{14} \frac{\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{14}} \]

- Electric field at point B:
\[ \vec{E}_B = \frac{kQ}{R_B^2} \hat{r}_B = \frac{kQ}{3} \frac{\vec{r}_B}{|\vec{r}_B|} = \frac{kQ}{3} \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \]

- Electric field at point C:
\[ \vec{E}_C = \frac{kQ}{R_C^2} \hat{r}_C = \frac{kQ}{12} \frac{\vec{r}_C}{|\vec{r}_C|} = \frac{kQ}{12} \frac{2\hat{i} + 2\hat{j} + 2\hat{k}}{\sqrt{12}} = \frac{kQ}{12} \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \]

### Step 4: Compare the magnitudes of the electric fields
Now we can compare the magnitudes of the electric fields:
- Magnitude of \(\vec{E}_A\):
\[ |\vec{E}_A| = \frac{kQ}{14\sqrt{14}} \]

- Magnitude of \(\vec{E}_B\):
\[ |\vec{E}_B| = \frac{kQ}{3\sqrt{3}} \]

- Magnitude of \(\vec{E}_C\):
\[ |\vec{E}_C| = \frac{kQ}{12\sqrt{3}} \]

### Step 5: Establish the relationship
To find the relationship between \(|\vec{E}_B|\) and \(|\vec{E}_C|\):
\[ |\vec{E}_B| = \frac{kQ}{3\sqrt{3}}, \quad |\vec{E}_C| = \frac{kQ}{12\sqrt{3}} \]
From this, we can see that:
\[ |\vec{E}_B| = 4 \cdot |\vec{E}_C| \]

### Conclusion
Thus, the electric field at point B is four times the electric field at point C:
\[ \vec{E}_B = 4 \cdot \vec{E}_C \]