Question

# The electric potential at a point $$(x, y)$$ in an electric field is given by $$V=6 x y+y^{2}-x^{2}$$ Calculate the electric field at that point.

Solution

## $$\begin{array}{c}\vec{E}=-\vec{\nabla} V, \vec{E}=-\dfrac{\partial V}{\partial x} \hat{i}-\dfrac{\partial V}{\partial y} \hat{j}-\dfrac{\partial V}{\partial z} \hat{k} \\\vec{V}=\left[6 x y+y^{2}-x^{2}\right] \\\vec{E}=-\dfrac{\partial}{\partial x}\left(6 x y+y^{2}-x^{2}\right) \hat{i}-\dfrac{\partial}{\partial y}\left[6 x y+y^{2}-x^{2}\right] \dot{j} -\dfrac{\partial}{\partial z}\left[6 x y+y^{2}-x^{2}\right] \hat{k}\\\vec{E}=-[6 y-2 x] \hat{i}-[6 x+2 y] \hat{j}-0 \\\vec{E}=(2 x-6 y) \hat{i}-(6 x+2 y) \hat{j \ } \mathrm{volt / m} \end{array}$$Physics

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