Question

Find the electric field for the given electrostatic potential field at point \(P(2,3)\): \[ V = 5(x^2 - y^2) \]

• A. \( -20\hat{i} + 30\hat{j} \)
• B. \( 20\hat{i} + 30\hat{j} \)
• C. \( 30\hat{i} - 20\hat{j} \)
• D. \( 30\hat{i} + 20\hat{j} \)

Answer 1

The Correct Option is A

Concept: The electric field is related to electrostatic potential by the negative gradient of potential: \[ \vec{E} = - \nabla V \] In Cartesian coordinates, \[ \vec{E} = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}\right) \] Thus, we first calculate the partial derivatives of the potential function. Step 1: {\color{red}Differentiate the potential with respect to \(x\) and \(y\).} Given \[ V = 5(x^2 - y^2) \] Partial derivatives: \[ \frac{\partial V}{\partial x} = 10x \] \[ \frac{\partial V}{\partial y} = -10y \] Step 2: {\color{red}Apply the negative gradient relation.} \[ \vec{E} = -(10x\hat{i} - 10y\hat{j}) \] \[ \vec{E} = -10x\hat{i} + 10y\hat{j} \] Step 3: {\color{red}Substitute the coordinates of point \(P(2,3)\).} \[ \vec{E} = -10(2)\hat{i} + 10(3)\hat{j} \] \[ \vec{E} = -20\hat{i} + 30\hat{j} \] Hence, the electric field at point \(P(2,3)\) is \[ \boxed{-20\hat{i} + 30\hat{j}} \]