The electric potential in a region of space is V=200/2x2+y2,where x and y are in meters. What are the strength and direction of the electric field at 1x, y2=12.0 m, 1.0 m2? Give the direction as an angle cw or ccw (specify which) from the positive x-axis

We have given that the electric potential at the region of space is$v=\frac{200}{\sqrt{x^2+y^2}}$.

The derivative of the expression for the potential with respect to each of the variables and the negative of these derivatives will give the formula for the components of the electric field.

E(x, y) = $\left(-\frac{dV(x,y)}{dx},-\frac{dV(x,y)}{dy}\right)$

So, calculate the derivative of an expression of such a form:

localid="1648482660213" $\frac{d}{d{u}_1}\frac{c}{\sqrt{{u_1}^2+{u_2}^2}}=-\frac{c}{2\sqrt{{{(u_1}^2+{u_2}^2)}^3}}\frac{d}{d{u}_1}({u_1}^2+{u_2}^2)=-\frac{c{u}_1}{\sqrt{{({u_1}^2+{u_2}^2)}^3}}$

Now, the electric field will be given by

$E\left(x,y\right)=\left(\frac{200x}{\sqrt{{(x^2+y^2)}^3}},\frac{200y}{\sqrt{{(x^2+y^2)}^3}}\right)$

point (2,1) this expression reveals

$E\left(2,1\right)=\left(\frac{200.2}{\sqrt{{(2^2+1^2)}^3}},\frac{200.1}{\sqrt{{(2^2+1^2)}^3}}\right)$

so components will be

$$\begin{gathered} Ez=35.{77}_{} \\ Ey=17.88 \end{gathered}$$

it means the magnitude will be

$\overline{)\stackrel{\rightharpoonup }{E}(2,1\overline{))}=\sqrt{{E_x}^2+{E_y}^2}=\sqrt{35.{77}^2+17.{88}^2}}=40$

the magnitude of the electric field will be 40 V/m

and we can also find the direction as

$\alpha ={\tan }^{-1}\left(\frac{Ey}{Ez}\right)={\tan }^{-1}\left(\frac{1}{2}\right)={27}^{°}$