Question

the electric potential at a pont is given by V=z(z^2-4x^2) Calculate the electric field E at that point​

Answer 1

Answer:

See below.

Explanation:

The electric potential is given by,

$V\left( x,z\right) =z\cdot \left( z^{2}-4x^{2}\right)$

We know the relation,

$\begin{aligned}\dfrac{\partial V}{\partial x}=E_{x}\\ \dfrac{ \partial V}{\partial z}=E_{z}\end{aligned}$

First, we differentiate V( x , z ) with respect to x treating z as a constant.

$\dfrac{\partial V}{\partial x}=z\left( -8x\right) =-8zx$

Next, we differentiate V( x , z ) with respect to z treating x as a constant.

$\begin{aligned}\dfrac{\partial V}{\partial z}=z\left( 2z\right) +\left( z^{2}-4x^{2}\right) \\ =3z^{2}-4x^{2}\end{aligned}$

Note that the electric potential is independent of the y coordinate and hence the change in electric potential with respect to the y coordinate is zero.

$\dfrac{\partial V}{\partial y}=0$

Now, writing the electric field E,

$\overrightarrow{E}=E_{x}\widehat{i}+E_y \widehat{j} + E_z \widehat{k}$

$= (-8zx) \widehat{i} + ( 3z^2 - 4x^2 ) \widehat{k}$