Electric potential at any point is `V = - 5 x + 3y + sqrt(15) z`, then the magnitude of the electric field is

Options

A. `3 sqrt(2)`
B. `4 sqrt(2)`
C. `5 sqrt(2)`
D. 7

Answer

To find the magnitude of the electric field given the electric potential \( V = -5x + 3y + \sqrt{15} z \), we will follow these steps:

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

### Step 2: Calculate the components of the electric field
The components of the electric field can be calculated using the following formulas:
- \( E_x = -\frac{\partial V}{\partial x} \)
- \( E_y = -\frac{\partial V}{\partial y} \)
- \( E_z = -\frac{\partial V}{\partial z} \)

#### Step 2.1: Calculate \( E_x \)
\[ E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(-5x + 3y + \sqrt{15} z) = -(-5) = 5 \]

#### Step 2.2: Calculate \( E_y \)
\[ E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(-5x + 3y + \sqrt{15} z) = -3 \]

#### Step 2.3: Calculate \( E_z \)
\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(-5x + 3y + \sqrt{15} z) = -\sqrt{15} \]

### Step 3: Calculate the magnitude of the electric field
The magnitude of the electric field \( E \) is given by:
\[ E = \sqrt{E_x^2 + E_y^2 + E_z^2} \]

Substituting the values we found:
\[ E = \sqrt{(5)^2 + (-3)^2 + (-\sqrt{15})^2} \]
\[ E = \sqrt{25 + 9 + 15} \]
\[ E = \sqrt{49} = 7 \]

### Conclusion
The magnitude of the electric field is \( E = 7 \).

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