Question

Which of the following represents the correct vector form of the electric field in three dimensions in terms of the potential gradient?

Options

• \[\vec{\mathbf{E}} = -\left(\frac{dx}{dV}\hat{i}+\frac{dy}{dV}\hat{j}+\frac{dz}{dV}\hat{k}\right)\]

• \[\vec{\mathbf{E}} = -\nabla^2 V = -\left(\frac{\partial^2 V}{\partial x^2}\hat{i}+\frac{\partial^2 V}{\partial y^2}\hat{j}+\frac{\partial^2 V}{\partial z^2}\hat{k}\right)\]

• \[\vec{\mathbf{E}} = \nabla V = \left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)\]

• \[\vec{\mathbf{E}} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)\]

Answer 1

In three dimensions, the electric field is the negative gradient of potential, defined as \[\vec{\mathbf{E}} = -\nabla V = -\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)\].