Let's solve the problem step by step as per Chegg guidelines. --- ### **Given:** - **Point:** \( P = (-3.0, 2.5, -18.0) \) m - **Potential:** \[ V(x, y, z) = \frac{x^2}{y^2 + 1} + \frac{y^2}{x^2 + y} + z \] --- ### **Step 1: Understanding the Electric Field and Potential** The **electric field** \( \vec{E} \) is related to the electric potential \( V \) by: \[ \vec{E} = -\nabla V \] where \( \nabla V \) is the gradient of \( V \). In Cartesian coordinates: \[ \vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \] --- ### **Step 2: Find the Partial Derivatives** #### **a) Partial derivative with respect to \( x \):** Let's compute \( \frac{\partial V}{\partial x} \): \[ V(x, y, z) = \frac{x^2}{y^2 + 1} + \frac{y^2}{x^2 + y} + z \] - The derivative of \( \frac{x^2}{y^2 + 1} \) with respect to \( x \) is \( \frac{2x}{y^2 + 1} \). - The derivative of \( \frac{y^2}{x^2 + y} \) with respect to \( x \) is: - Use quotient rule: \( \frac{d}{dx} \left( \frac{y^2}{x^2 + y} \right) = y^2 \cdot \frac{d}{dx} \left( \frac{1}{x^2 + y} \right) \) - \( \frac{d}{dx} \left( \frac{1}{x^2 + y} \right) = -\frac{2x}{(x^2 + y)^2} \) - So, \( - \frac{2x y^2}{(x^2 + y)^2} \) - The derivative of \( z \) with respect to \( x \) is zero. So: \[ \frac{\partial V}{\partial x} = \frac{2x}{y^2 + 1} - \frac{2x y^2}{(x^2 + y)^2} \] --- #### **b) Partial derivative with respect to \( y \):** - The derivative of \( \frac{x^2}{y^2 + 1} \) with respect to \( y \) is: - \( \frac{d}{dy} \left( \frac{x^2}{y^2 + 1} \right) = x^2 \cdot \frac{d}{dy} \left( \frac{1}{y^2 + 1} \right) = -\frac{2x^2 y}{(y^2 + 1)^2} \) - The derivative of \( \frac{y^2}{x^2 + y} \) with respect to \( y \): - Use quotient rule: \( \frac{d}{dy} \left( \frac{y^2}{x^2 + y} \right) = \frac{2y (x^2 + y) - y^2 \cdot 1}{(x^2 + y)^2} \) - Simplify numerator: \( 2y(x^2 + y) - y^2 = 2x^2 y + 2y^2 - y^2 = 2x^2 y + y^2 \) - So, \( \frac{2x^2 y + y^2}{(x^2 + y)^2} \) - The derivative of \( z \) with respect to \( y \) is zero. So: \[ \frac{\partial V}{\partial y} = -\frac{2x^2 y}{(y^2 + 1)^2} + \frac{2x^2 y + y^2}{(x^2 + y)^2} \] --- #### **c) Partial derivative with respect to \( z \):** - The derivative of \( V \) with respect to \( z \) is simply 1. \[ \frac{\partial V}{\partial z} = 1 \] --- ### **Step 3: Substitute the Point \( (-3.0, 2.5, -18.0) \) into the Derivatives** Let \( x = -3.0 \), \( y = 2.5 \), \( z = -18.0 \). #### **Calculate all terms step by step:** ##### **a) \( \frac{\partial V}{\partial x} \):** - \( y^2 + 1 = (2.5)^2 + 1 = 6.25 + 1 = 7.25 \) - \( x^2 + y = (-3.0)^2 + 2.5 = 9 + 2.5 = 11.5 \) - \( (x^2 + y)^2 = (11.5)^2 = 132.25 \) - \( y^2 = (2.5)^2 = 6.25 \) - Plug in values: \[ \frac{\partial V}{\partial x} = \frac{2 \cdot (-3.0)}{7.25} - \frac{2 \cdot (-3.0) \cdot 6.25}{132.25} \] \[ = \frac{-6.0}{7.25} - \frac{-37.5}{132.25} \] \[ = -0.8276 + 0.2836 \] \[ = -0.544 \] ##### **b) \( \frac{\partial V}{\partial y} \):** - \( x^2 = 9 \) - \( (y^2 + 1)^2 = (7.25)^2 = 52.5625 \) - \( 2x^2 y = 2 \cdot 9 \cdot 2.5 = 45 \) - \( (x^2 + y)^2 = 132.25 \) - \( y^2 = 6.25 \) First term: \[ -\frac{2x^2 y}{(y^2 + 1)^2} = -\frac{45}{52.5625} = -0.857 \] Second term numerator: \[ 2x^2 y + y^2 = 45 + 6.25 = 51.25 \] \[ \frac{51.25}{132.25} = 0.3877 \] So, \[ \frac{\partial V}{\partial y} = -0.857 + 0.388 = -0.469 \] ##### **c) \( \frac{\partial V}{\partial z} = 1 \)** --- ### **Step 4: Electric Field Components** \[ E_x = -\frac{\partial V}{\partial x} = -(-0.544) = 0.544 \text{ V/m} \] \[ E_y = -\frac{\partial V}{\partial y} = -(-0.469) = 0.469 \text{ V/m} \] \[ E_z = -\frac{\partial V}{\partial z} = -1 \text{ V/m} \] --- ### **Step 5: Magnitude of the Electric Field** \[ |\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \] \[ = \sqrt{(0.544)^2 + (0.469)^2 + (1)^2} \] \[ = \sqrt{0.296 + 0.220 + 1} \] \[ = \sqrt{1.516} \] \[ = 1.23 \text{ V/m} \] --- ## **Final Answer** \[ \boxed{1.2 \text{ V/m}} \] (to two significant figures, as requested) --- ### **Summary of Steps** 1. **Found the gradient of the potential \( V \) with respect to each variable.** 2. **Substituted the coordinates into the partial derivatives.** 3. **Calculated the components of the electric field.** 4. **Found the magnitude of the electric field vector.** 5. **Expressed the answer with appropriate units and significant figures.** If you need more explanation on any step, let me know!

Certainly! Here's a detailed explanation of each step in words, following the process to find the electric field at the given point from the potential function: --- ### **Understanding the Problem** We are asked to find the electric field at a specific point in space, given a potential function \( V(x, y, z) \). The electric field is related to the potential via the negative gradient: \[ \vec{E} = -\nabla V \] which means that to find the electric field components, we need to compute the partial derivatives of \( V \) with respect to each coordinate, then take their negatives. --- ### **Step 1: Write down the potential function** The potential function provided is: \[ V(x, y, z) = \frac{x^2}{y^2 + 1} + \frac{y^2}{x^2 + y} + z \] This function combines rational expressions involving \( x \) and \( y \), plus a linear term in \( z \). --- ### **Step 2: Find the partial derivatives of \( V \)** To compute the electric field components, we need: \[ E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y}, \quad E_z = -\frac{\partial V}{\partial z} \] Let's examine these derivatives: **a) Derivative with respect to \( x \):** - The first term, \( \frac{x^2}{y^2 + 1} \), is straightforward: its derivative with respect to \( x \) is \( \frac{2x}{y^2 + 1} \). - The second term, \( \frac{y^2}{x^2 + y} \), involves \( x \) in the denominator. Using the quotient rule, its derivative with respect to \( x \) is: \[ -\frac{2x y^2}{(x^2 + y)^2} \] - The \( z \) term does not depend on \( x \), so its derivative is zero. **b) Derivative with respect to \( y \):** - For \( \frac{x^2}{y^2 + 1} \), the derivative is: \[ -\frac{2x^2 y}{(y^2 + 1)^2} \] - For \( \frac{y^2}{x^2 + y} \), use the quotient rule: Numerator derivative: \( 2y \) Denominator: \( x^2 + y \) Derivative of the second term: \[ \frac{(2y)(x^2 + y) - y^2 \cdot 1}{(x^2 + y)^2} = \frac{2y x^2 + 2 y^2 - y^2}{(x^2 + y)^2} = \frac{2 y x^2 + y^2}{(x^2 + y)^2} \] - The \( z \) term again contributes zero. **c) Derivative with respect to \( z \):** Since only the linear \( z \) term depends on \( z \), its derivative is simply 1. --- ### **Step 3: Substitute the point values into derivatives** Given point: \( x = -3.0 \), \( y = 2.5 \), \( z = -18.0 \). - Compute necessary quantities: \[ y^2 + 1 = 6.25 + 1 = 7.25 \] \[ x^2 + y = 9 + 2.5 = 11.5 \] \[ (x^2 + y)^2 = (11.5)^2 = 132.25 \] \[ y^2 = 6.25 \] - **Calculate \( \frac{\partial V}{\partial x} \):** \[ \frac{2x}{y^2 + 1} = \frac{2 \times -3}{7.25} = -\frac{6}{7.25} \approx -0.8276 \] \[ -\frac{2x y^2}{(x^2 + y)^2} = -\frac{2 \times -3 \times 6.25}{132.25} = -\frac{-37.5}{132.25} \approx +0.2836 \] Adding these: \[ \frac{\partial V}{\partial x} \approx -0.8276 + 0.2836 = -0.544 \] - **Calculate \( \frac{\partial V}{\partial y} \):** \[ -\frac{2x^2 y}{(y^2 + 1)^2} = -\frac{2 \times 9 \times 2.5}{(7.25)^2} = -\frac{45}{52.5625} \approx -0.857 \] \[ \frac{2 y x^2 + y^2}{(x^2 + y)^2} = \frac{2 \times 2.5 \times 9 + 6.25}{132.25} = \frac{45 + 6.25}{132.25} \approx 0.388 \] Adding: \[ \frac{\partial V}{\partial y} \approx -0.857 + 0.388 = -0.469 \] - **Calculate \( \frac{\partial V}{\partial z} \):** \[ = 1 \] --- ### **Step 4: Find the electric field components** Recall: \[ \vec{E} = -\nabla V \] Thus: \[ E_x = -(-0.544) = 0.544 \, \text{V/m} \] \[ E_y = -(-0.469) = 0.469 \, \text{V/m} \] \[ E_z = -1 = -1 \, \text{V/m} \] --- ### **Step 5: Calculate the magnitude of the electric field** The magnitude is: \[ |\vec{E}| = \sqrt{(0.544)^2 + (0.469)^2 + (-1)^2} \] which approximately equals: \[ \sqrt{0.296 + 0.220 + 1} = \sqrt{1.516} \approx 1.23 \, \text{V/m} \] --- ### **Final Result:** **The magnitude of the electric field at the point \((-3.0, 2.5, -18.0)\) m is approximately \( \boxed{1.2\, \text{V/m}} \).** --- This step-by-step explanation connects the mathematical derivatives to the physical concept of the electric field, illustrating how the potential function determines the electric field vector in space.