Let's break down and solve the problem step-by-step. --- ## **Problem Restatement** The **electric potential** in a certain region of space is given by: \[ V(x, y) = Ax + By \] where: - \(A = 9.00\, \text{V/m}\) - \(B = -8.00\, \text{V/m}\) You are asked for **Part A**: **Calculate the magnitude of the electric field at the point in the region with coordinates \(x = 3.00\, \text{m}, y = 4.00\, \text{m}\).** --- ## **Step 1: Relate Electric Field to Potential** The **electric field** \(\vec{E}\) is related to the potential \(V(x, y)\) by: \[ \vec{E} = -\nabla V \] For \(V(x, y) = Ax + By\): \[ \frac{\partial V}{\partial x} = A \\ \frac{\partial V}{\partial y} = B \] So, \[ \vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} \right) = -(A \hat{i} + B \hat{j}) \] \[ \vec{E} = -A\hat{i} - B\hat{j} \] --- ## **Step 2: Plug in the Values** Given \(A = 9.00\, \text{V/m}\), \(B = -8.00\, \text{V/m}\): \[ \vec{E} = -(9.00\, \hat{i} + (-8.00)\, \hat{j}) \text{ V/m} \] \[ \vec{E} = -9.00\, \hat{i} + 8.00\, \hat{j} \text{ V/m} \] --- ## **Step 3: Find the Magnitude of the Electric Field** \[ |\vec{E}| = \sqrt{(-9.00)^2 + (8.00)^2} \] \[ |\vec{E}| = \sqrt{81 + 64} \] \[ |\vec{E}| = \sqrt{145} \] \[ |\vec{E}| \approx 12.\, \text{V/m} \] --- ## **Step 4: Final Answer** \[ \boxed{|\vec{E}| = 12.\, \text{V/m}} \] --- ### **Explanation** - The electric field in a region where potential is linear in \(x\) and \(y\) is constant in magnitude and direction. - The magnitude does **not** depend on the specific point (x, y). --- If you need the direction (angle), let me know!