Electrical potential `V in space as a function of `co-ordinates is given by `,V=(1)/(x)+(1)/(y)+(1)/(z)`
then the electric field intensity at `(1,1,1)` is given by:

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))`

Explanation

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 will 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 three dimensions 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
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 components of the electric field
Now we can express the electric field components:
\[ \mathbf{E} = -\left( \frac{\partial V}{\partial x} \mathbf{i} + \frac{\partial V}{\partial y} \mathbf{j} + \frac{\partial V}{\partial z} \mathbf{k} \right) \]
Substituting the values we computed:
\[ \mathbf{E} = -\left( -\frac{1}{x^2} \mathbf{i} - \frac{1}{y^2} \mathbf{j} - \frac{1}{z^2} \mathbf{k} \right) \]
\[ \mathbf{E} = \left( \frac{1}{x^2} \mathbf{i} + \frac{1}{y^2} \mathbf{j} + \frac{1}{z^2} \mathbf{k} \right) \]

### Step 5: Evaluate the electric field at the point (1, 1, 1)
Substituting \( x = 1 \), \( y = 1 \), and \( z = 1 \):
\[ \mathbf{E} = \left( \frac{1}{1^2} \mathbf{i} + \frac{1}{1^2} \mathbf{j} + \frac{1}{1^2} \mathbf{k} \right) \]
\[ \mathbf{E} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]

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

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