Electric potential V in space as a function of cartesian co-ordinates is given by `V=(1)/(x)+(1)/(y) +(1)/(z)`. Then find electric field intensity at (1, 1, 1)

Options

A. `(hat(i) + hat(j) + hat(k))`
B. `hat(i) + hat (J) + hat(K)`
C. zero
D. `(1)/(sqrt(3))(hat(i)+hat(j)+hat(k))`

Answer

To find the electric field intensity at the point (1, 1, 1) given the electric potential \( V = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \), we can follow these steps:

### Step 1: Understand the relationship between electric potential and electric field
The electric field \( \mathbf{E} \) is related to the electric potential \( V \) by the equation:
\[ \mathbf{E} = -\nabla V \]
where \( \nabla V \) is the gradient of the potential.

### Step 2: Calculate the gradient of the potential
The gradient in Cartesian coordinates is given by:
\[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \]

### Step 3: Compute the partial derivatives
Given \( V = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \), we will compute the partial derivatives:

1. **Partial derivative with respect to \( x \)**:
\[ \frac{\partial V}{\partial x} = -\frac{1}{x^2} \]

2. **Partial derivative with respect to \( y \)**:
\[ \frac{\partial V}{\partial y} = -\frac{1}{y^2} \]

3. **Partial derivative with respect to \( z \)**:
\[ \frac{\partial V}{\partial z} = -\frac{1}{z^2} \]

### Step 4: Write the gradient vector
Now, substituting the partial derivatives into the gradient:
\[ \nabla V = \left( -\frac{1}{x^2}, -\frac{1}{y^2}, -\frac{1}{z^2} \right) \]

### Step 5: Calculate the electric field
Using the relationship \( \mathbf{E} = -\nabla V \):
\[ \mathbf{E} = -\left( -\frac{1}{x^2}, -\frac{1}{y^2}, -\frac{1}{z^2} \right) = \left( \frac{1}{x^2}, \frac{1}{y^2}, \frac{1}{z^2} \right) \]

### Step 6: Substitute the coordinates (1, 1, 1)
Now we substitute \( x = 1 \), \( y = 1 \), and \( z = 1 \):
\[ \mathbf{E} = \left( \frac{1}{1^2}, \frac{1}{1^2}, \frac{1}{1^2} \right) = (1, 1, 1) \]

### Final Result
Thus, the electric field intensity at the point (1, 1, 1) is:
\[ \mathbf{E} = \hat{i} + \hat{j} + \hat{k} \]