The electric potential at a space point P (X, Y, Z) is given as `V= x^(2)+y^(2)+z^(2)`. The modulus of the electric field at that point is proportional to

Options

A. `V^((1)/(2))`
B. V
C. `V^((3)/(2))`
D. `V^(2)`

Answer

To solve the problem, we need to find the relationship between the electric potential \( V \) and the electric field \( \vec{E} \) at a point \( P(X, Y, Z) \). The electric potential is given by:

\[ V = x^2 + y^2 + z^2 \]

### Step 1: Calculate the Electric Field
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the negative gradient of the potential:

\[ \vec{E} = -\nabla V \]

Where \( \nabla V \) is the gradient of the potential \( V \). In Cartesian coordinates, the gradient is given by:

\[ \nabla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \right) \]

### Step 2: Compute the Partial Derivatives
Now we compute the partial derivatives of \( V \):

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

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

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

### Step 3: Write the Electric Field Vector
Now we can write the electric field vector \( \vec{E} \):

\[ \vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) = -\left( 2x \hat{i} + 2y \hat{j} + 2z \hat{k} \right) \]

Thus, we have:

\[ \vec{E} = -2x \hat{i} - 2y \hat{j} - 2z \hat{k} \]

### Step 4: Calculate the Magnitude of the Electric Field
The magnitude of the electric field \( |\vec{E}| \) is given by:

\[ |\vec{E}| = \sqrt{(-2x)^2 + (-2y)^2 + (-2z)^2} \]

Calculating this gives:

\[ |\vec{E}| = \sqrt{4x^2 + 4y^2 + 4z^2} = 2\sqrt{x^2 + y^2 + z^2} \]

### Step 5: Relate the Magnitude of the Electric Field to the Potential
We know that:

\[ V = x^2 + y^2 + z^2 \]

Thus, we can express the magnitude of the electric field in terms of the potential:

\[ |\vec{E}| = 2\sqrt{V} \]

This shows that the modulus of the electric field \( |\vec{E}| \) is proportional to \( \sqrt{V} \).

### Final Answer
The modulus of the electric field at that point is proportional to \( \sqrt{V} \).

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