Section 3.5: Relation E = -∇V
The electric field is related to the electric potential by the gradient relation:
Mathematical Expression:
\[ \vec{E} = - \nabla V \]
This means the electric field points in the direction of maximum decrease of the potential and has a magnitude equal to the rate of decrease per unit distance.
Example 1
Given a potential function \( V(x,y,z) = k x^2 y \), find the electric field \( \vec{E} \).
\[ \vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) = - (2 k x y \hat{i} + k x^2 \hat{j} + 0 \hat{k}) = -2kxy \hat{i} - kx^2 \hat{j} \]
Practice Problems
- Compute the electric field from \( V = A/r \) for a point charge at the origin.
- Show that for a uniform field along x-axis, \( V = -Ex \), the field is recovered using \( \vec{E} = -\nabla V \).
- Given \( V(x,y) = 3x^2 + 2y^2 \), find \( \vec{E} \) at point (1,1).
- Explain why electric field lines are always perpendicular to equipotential surfaces.
- Verify \( \vec{E} = -\nabla V \) for \( V = k(x^2 + y^2 + z^2) \).