A charge `q = (-2 xx 10^(-9))` is placed at (1 m, 2m, 3m). There is a point `P = (2m , -3m, 4 m)`
A unit vector in the direction of electric field at P due to charge q will be

Options

A. `(1)/(3 sqrt3) (hat(i) - 5 hat(j) + hat(k))`
B. `(1)/(3 sqrt3) ( - hat(i) + 5 hat(j) - hat(k))`
C. `(1)/(5 sqrt2) (3 hat(i) - 4 hat(j) + 5 hat(k))`
D. `(1)/(5 sqrt2) ( - 3 hat(i) + 4 hat(j) - 5 hat(k))`

Answer

To find the unit vector in the direction of the electric field at point P due to the charge \( q \), we will follow these steps:

### Step 1: Identify the position vectors
Let the position of the charge \( q \) be \( \mathbf{A} = (1, 2, 3) \) and the position of point \( P \) be \( \mathbf{P} = (2, -3, 4) \).

### Step 2: Calculate the position vector from P to A
The position vector \( \mathbf{r} \) from point \( P \) to charge \( q \) at point \( A \) is given by:
\[ \mathbf{r} = \mathbf{A} - \mathbf{P} = (1, 2, 3) - (2, -3, 4) \]
Calculating this gives:
\[ \mathbf{r} = (1 - 2, 2 - (-3), 3 - 4) = (-1, 5, -1) \]

### Step 3: Calculate the magnitude of the position vector
The magnitude \( |\mathbf{r}| \) is calculated using the formula:
\[ |\mathbf{r}| = \sqrt{(-1)^2 + 5^2 + (-1)^2} = \sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3} \]

### Step 4: Calculate the unit vector
The unit vector \( \hat{\mathbf{r}} \) in the direction of the electric field at point \( P \) due to charge \( q \) is given by:
\[ \hat{\mathbf{r}} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{(-1, 5, -1)}{3\sqrt{3}} \]
This simplifies to:
\[ \hat{\mathbf{r}} = \left(-\frac{1}{3\sqrt{3}}, \frac{5}{3\sqrt{3}}, -\frac{1}{3\sqrt{3}}\right) \]

### Step 5: Conclusion
Thus, the unit vector in the direction of the electric field at point \( P \) due to charge \( q \) is:
\[ \hat{\mathbf{E}} = \left(-\frac{1}{3\sqrt{3}}, \frac{5}{3\sqrt{3}}, -\frac{1}{3\sqrt{3}}\right) \]

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