A point charge `q=100 mu C` is located in the `x-y` plane at the point with position vector `vec(r)_(0)=2 hat(i)+3 hat(k)+hat(k) .` What is the magnitude of electric field vector (in `(kV) / (m)` ) at the point with position vector `vec(r)=8 hat(i)-5 hat(j)+k ?`

Answer

To find the magnitude of the electric field vector at the given point due to a point charge, we can follow these steps:

### Step 1: Identify the position vectors
The position vector of the point charge \( \vec{r}_0 \) is given as:
\[ \vec{r}_0 = 2\hat{i} + 3\hat{j} + 1\hat{k} \]
The position vector of the point where we want to find the electric field \( \vec{r} \) is:
\[ \vec{r} = 8\hat{i} - 5\hat{j} + 1\hat{k} \]

### Step 2: Calculate the displacement vector \( \Delta \vec{r} \)
The displacement vector from the charge to the point of interest is given by:
\[ \Delta \vec{r} = \vec{r} - \vec{r}_0 = (8\hat{i} - 5\hat{j} + 1\hat{k}) - (2\hat{i} + 3\hat{j} + 1\hat{k}) \]
Calculating this gives:
\[ \Delta \vec{r} = (8 - 2)\hat{i} + (-5 - 3)\hat{j} + (1 - 1)\hat{k} = 6\hat{i} - 8\hat{j} + 0\hat{k} \]

### Step 3: Calculate the magnitude of the displacement vector
The magnitude of the displacement vector \( r \) is calculated as:
\[ r = |\Delta \vec{r}| = \sqrt{(6)^2 + (-8)^2 + (0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m} \]

### Step 4: Use the formula for the electric field due to a point charge
The electric field \( \vec{E} \) due to a point charge \( q \) at a distance \( r \) is given by:
\[ \vec{E} = k \frac{q}{r^2} \hat{r} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant, \( q = 100 \, \mu C = 100 \times 10^{-6} \, C \), and \( \hat{r} \) is the unit vector in the direction of \( \Delta \vec{r} \).

### Step 5: Calculate the unit vector \( \hat{r} \)
The unit vector \( \hat{r} \) is given by:
\[ \hat{r} = \frac{\Delta \vec{r}}{|\Delta \vec{r}|} = \frac{6\hat{i} - 8\hat{j}}{10} = 0.6\hat{i} - 0.8\hat{j} \]

### Step 6: Substitute values into the electric field formula
Now substituting the values into the electric field formula:
\[ \vec{E} = k \frac{q}{r^2} \hat{r} = 9 \times 10^9 \frac{100 \times 10^{-6}}{10^2} (0.6\hat{i} - 0.8\hat{j}) \]
Calculating this gives:
\[ \vec{E} = 9 \times 10^9 \frac{100 \times 10^{-6}}{100} (0.6\hat{i} - 0.8\hat{j}) = 9 \times 10^9 \times 10^{-6} (0.6\hat{i} - 0.8\hat{j}) = 9 \times 0.6 \hat{i} - 9 \times 0.8 \hat{j} \]
\[ \vec{E} = 5.4 \hat{i} - 7.2 \hat{j} \, \text{kV/m} \]

### Step 7: Convert to kV/m
To express the result in kV/m, we can directly state:
\[ \vec{E} = 5.4 \hat{i} - 7.2 \hat{j} \, \text{kV/m} \]

### Final Result
The magnitude of the electric field vector at the given point is:
\[ \vec{E} = 5.4 \hat{i} - 7.2 \hat{j} \, \text{kV/m} \]