Question

The electric potential for various points in $x$-$y$ plane is given by $V = 1.0 x^2 - 2.0 y^2$, where $V$ is in volts and $x$, $y$ are in metres. The angle that the electric field at point $(2.0$ m$, 1.0$ m$)$ makes with the positive $x$-axis is:

• A. $45^\circ$
• B. $90^\circ$
• C. $135^\circ$
• D. $315^\circ$

Hint:

Use \(\vec{E} = -\nabla V\) to get \(E_x = -2x\) and \(E_y = 4y\). At (2, 1) this gives \((-4, 4)\), which is at \(135^{\circ}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The electric field is the negative gradient of the potential. In two dimensions, \(E_x = -\dfrac{\partial V}{\partial x}\) and \(E_y = -\dfrac{\partial V}{\partial y}\).
The direction of the field is then found from the signs and sizes of \(E_x\) and \(E_y\).

Step 2: Differentiate the Potential:
Given \(V = 1.0x^2 - 2.0y^2\).
\[ E_x = -\frac{\partial V}{\partial x} = -2.0x \]
\[ E_y = -\frac{\partial V}{\partial y} = +4.0y \]

Step 3: Put in the Point (2.0 m, 1.0 m):
\[ E_x = -2.0 \times 2.0 = -4.0 \text{ V/m} \]
\[ E_y = 4.0 \times 1.0 = +4.0 \text{ V/m} \]
So the field points left (negative x) and up (positive y). Both components have equal size.

Step 4: Find the Angle:
\[ \tan\theta = \frac{E_y}{E_x} = \frac{4.0}{-4.0} = -1 \]
The reference angle is \(45^{\circ}\). The vector lies in the second quadrant (x negative, y positive).
So the angle with the positive x-axis is \(180^{\circ} - 45^{\circ} = 135^{\circ}\).

Step 5: Why the Other Options Are Wrong:
\(45^{\circ}\) needs both components positive, but \(E_x\) is negative. \(90^{\circ}\) needs \(E_x = 0\), but \(E_x = -4.0\). \(315^{\circ}\) needs \(E_x\) positive and \(E_y\) negative, which is the opposite of what we found.

Final Answer:
The field at the point makes \(135^{\circ}\) with the positive x-axis. This is option (C). \[ \boxed{135^{\circ}} \]