At a distance r from a point located at origin in space, the electric potential varies as `V= 10r`. Find the electric field at `vecr = 3hati + 4 hatj - 5hatk`.

Options

A. `(sqrt2)(3hati+4hatj-5hatk)`
B. `(-sqrt2)(3hati+4hatj-5hatk)`
C. `(-sqrt3)(3hati+4hatj-5hatk)`
D. none of these

Explanation

To solve the problem, we need to find the electric field \( \vec{E} \) at the point given by the vector \( \vec{r} = 3\hat{i} + 4\hat{j} - 5\hat{k} \) when the electric potential \( V \) varies as \( V = 10r \), where \( r \) is the distance from the origin.

### Step 1: Express \( r \) in terms of coordinates
The distance \( r \) from the origin to the point \( (x, y, z) \) is given by:
\[ r = \sqrt{x^2 + y^2 + z^2} \]
For our point \( \vec{r} = 3\hat{i} + 4\hat{j} - 5\hat{k} \):
- \( x = 3 \)
- \( y = 4 \)
- \( z = -5 \)

Calculating \( r \):
\[ r = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} \]

### Step 2: Write the potential \( V \)
Substituting \( r \) into the potential equation:
\[ V = 10r = 10 \sqrt{50} = 10 \times 5\sqrt{2} = 50\sqrt{2} \]

### Step 3: Calculate the electric field \( \vec{E} \)
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the equation:
\[ \vec{E} = -\nabla V \]
In Cartesian coordinates, this can be expressed as:
\[ \vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]

### Step 4: Calculate the partial derivatives
First, we need to express \( V \) in terms of \( x, y, z \):
\[ V = 10 \sqrt{x^2 + y^2 + z^2} \]
Now, we compute the partial derivatives:

1. **For \( \frac{\partial V}{\partial x} \)**:
\[ \frac{\partial V}{\partial x} = 10 \cdot \frac{x}{\sqrt{x^2 + y^2 + z^2}} = 10 \cdot \frac{x}{r} \]

2. **For \( \frac{\partial V}{\partial y} \)**:
\[ \frac{\partial V}{\partial y} = 10 \cdot \frac{y}{\sqrt{x^2 + y^2 + z^2}} = 10 \cdot \frac{y}{r} \]

3. **For \( \frac{\partial V}{\partial z} \)**:
\[ \frac{\partial V}{\partial z} = 10 \cdot \frac{z}{\sqrt{x^2 + y^2 + z^2}} = 10 \cdot \frac{z}{r} \]

### Step 5: Substitute values of \( x, y, z \)
Now substituting \( x = 3 \), \( y = 4 \), and \( z = -5 \):
- \( r = \sqrt{50} \)

Calculating the components of \( \vec{E} \):
1. **For \( E_x \)**:
\[ E_x = -\frac{\partial V}{\partial x} = -10 \cdot \frac{3}{\sqrt{50}} = -\frac{30}{\sqrt{50}} = -\frac{30}{5\sqrt{2}} = -\frac{6}{\sqrt{2}} = -3\sqrt{2} \]

2. **For \( E_y \)**:
\[ E_y = -\frac{\partial V}{\partial y} = -10 \cdot \frac{4}{\sqrt{50}} = -\frac{40}{\sqrt{50}} = -\frac{40}{5\sqrt{2}} = -\frac{8}{\sqrt{2}} = -4\sqrt{2} \]

3. **For \( E_z \)**:
\[ E_z = -\frac{\partial V}{\partial z} = -10 \cdot \frac{-5}{\sqrt{50}} = \frac{50}{\sqrt{50}} = \frac{50}{5\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]

### Step 6: Write the electric field vector
Combining the components, we get:
\[ \vec{E} = -3\sqrt{2} \hat{i} - 4\sqrt{2} \hat{j} + 5\sqrt{2} \hat{k} \]

### Final Answer
\[ \vec{E} = -3\sqrt{2} \hat{i} - 4\sqrt{2} \hat{j} + 5\sqrt{2} \hat{k} \]