A particle in a certain conservative force field has a potential energy given by `V=(20xy)/z`. The force exerted on it is

Options

A. `((20y)/z)hati+((20x)/z)hatj+((20xy)/z^2)hatk`
B. `-((20y)/z)hati-((20x)/z)hatj+((20xy)/z^2)hatk`
C. `-((20y)/z)hati-((20x)/z)hatj-((20xy)/z^2)hatk`
D. `((20y)/z)hati+((20x)/z)hatj-((20xy)/z^2)hatk`

Explanation

To find the force exerted on a particle in a conservative force field with a given potential energy, we can use the relationship between force and potential energy. The force \(\mathbf{F}\) is related to the potential energy \(V\) by the equation:

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

Where \(\nabla V\) is the gradient of the potential energy. The gradient in three dimensions is given by:

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

Given the potential energy function:

\[ V = \frac{20xy}{z} \]

We will calculate the partial derivatives with respect to \(x\), \(y\), and \(z\).

### Step 1: Calculate \(\frac{\partial V}{\partial x}\)

\[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x} \left( \frac{20xy}{z} \right) = \frac{20y}{z} \]

### Step 2: Calculate \(\frac{\partial V}{\partial y}\)

\[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y} \left( \frac{20xy}{z} \right) = \frac{20x}{z} \]

### Step 3: Calculate \(\frac{\partial V}{\partial z}\)

Using the quotient rule for differentiation:

\[ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z} \left( \frac{20xy}{z} \right) = -\frac{20xy}{z^2} \]

### Step 4: Combine the results to find the force

Now we can express the force vector:

\[ \mathbf{F} = -\nabla V = -\left( \frac{20y}{z}, \frac{20x}{z}, -\frac{20xy}{z^2} \right) \]

This simplifies to:

\[ \mathbf{F} = \left( -\frac{20y}{z}, -\frac{20x}{z}, \frac{20xy}{z^2} \right) \]

### Final Answer

Thus, the force exerted on the particle is:

\[ \mathbf{F} = -\frac{20y}{z} \hat{i} - \frac{20x}{z} \hat{j} + \frac{20xy}{z^2} \hat{k} \]