Questions: What is the magnitude of the electric field at the point (6.60 i - 5.20 j + 4.20 k) m if the electric potential is given by V = 8.80 xyz^2, where V is in volts and x, y, and z are in meters?

Solution Steps

The electric field \(\mathbf{E}\) is related to the electric potential \(V\) by the negative gradient:

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

Given the potential \(V = 8.80 x y z^2\), we need to find the partial derivatives with respect to \(x\), \(y\), and \(z\).

1. Partial derivative with respect to \(x\):

   \[ \frac{\partial V}{\partial x} = 8.80 y z^2 \]

2. Partial derivative with respect to \(y\):

   \[ \frac{\partial V}{\partial y} = 8.80 x z^2 \]

3. Partial derivative with respect to \(z\):

   \[ \frac{\partial V}{\partial z} = 2 \times 8.80 x y z = 17.60 x y z \]

The point is \((x, y, z) = (6.60, -5.20, 4.20)\).

1. Electric field component \(E_x\):

   \[ E_x = -\frac{\partial V}{\partial x} = -8.80 \times (-5.20) \times (4.20)^2 \]

   \[ E_x = -8.80 \times (-5.20) \times 17.64 = 805.1712 \, \text{V/m} \]

2. Electric field component \(E_y\):

   \[ E_y = -\frac{\partial V}{\partial y} = -8.80 \times 6.60 \times (4.20)^2 \]

   \[ E_y = -8.80 \times 6.60 \times 17.64 = -1022.2224 \, \text{V/m} \]

3. Electric field component \(E_z\):

   \[ E_z = -\frac{\partial V}{\partial z} = -17.60 \times 6.60 \times (-5.20) \times 4.20 \]

   \[ E_z = -17.60 \times 6.60 \times (-5.20) \times 4.20 = 2520.1920 \, \text{V/m} \]

The magnitude of the electric field \(\mathbf{E}\) is given by:

\[ |\mathbf{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \]

Substituting the values:

\[ |\mathbf{E}| = \sqrt{(805.1712)^2 + (-1022.2224)^2 + (2520.1920)^2} \]

\[ |\mathbf{E}| = \sqrt{648297.0294 + 1044944.0495 + 6353093.6064} \]

\[ |\mathbf{E}| = \sqrt{8046334.6853} \approx 2836.7543 \, \text{V/m} \]

Final Answer