Question

Find electric field, for given electrostatic potential at 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. \(30\hat{i} + 20\hat{j}\)

Hint:

To find electric field from potential, always use \( \mathbf{E} = - \nabla V \). Differentiate the potential with respect to each coordinate and then apply the negative sign carefully.

Answer 1

The Correct Option is A

Step 1: Use the relation between electric field and potential.
The electric field is given by the negative gradient of electrostatic potential:
\[ \mathbf{E} = - \frac{\partial V}{\partial x} \hat{i} - \frac{\partial V}{\partial y} \hat{j} \]
Step 2: Differentiate the given potential.
The potential is given as:
\[ V = 5(x^2 - y^2) \] Now, calculate the partial derivatives with respect to \(x\) and \(y\):
- Partial derivative with respect to \(x\):
\[ \frac{\partial V}{\partial x} = 10x \] - Partial derivative with respect to \(y\):
\[ \frac{\partial V}{\partial y} = -10y \]
Step 3: Electric field components.
Thus, the electric field components are:
\[ \mathbf{E} = - (10x \hat{i}) + (10y \hat{j}) \]
Step 4: Substituting the coordinates of point \(P(2,3)\).
At point \(P(2,3)\), \( x = 2 \) and \( y = 3 \), so:
\[ \mathbf{E} = - (10(B) \hat{i}) + (10(C) \hat{j}) = - 20 \hat{i} + 30 \hat{j} \] Thus, the electric field at point \(P(2,3)\) is:
\[ \mathbf{E} = -20 \hat{i} + 30 \hat{j} \] Final Answer: (A) \(-20 \hat{i} + 30 \hat{j}\)