Question

The electric potential in a region of space is $V=(150x^2-200y^2)V$, where $x$and $y$are in meters. What are the strength and direction of the electric field at $(x,y)=(2.0m,2.0m)$? Give the direction as an angle $cw$or $ccw$(specify which) from the positive $x$-axis

Short answer

The strength of the electric field is$1000V/m$, pointing $53$degrees below the $+x$axis (clockwise).

Step-by-step solution

Given information 

We have given that the electric potential in a region of space is $V=1150x2$-$200y22V$, where $x$and $y$are in meters.

We need to find that the strength and direction of the electric field at $1x$, $y2=0.2m$, $2.0m2$

Simplify

Considering the potential

$V(x,y)=1500x^2-200y^2$,

At point $(2,2)$, the potential will be

$V(2,2)=150·2^2-200·2^2=2^2·50(3-4)=-200V$

For the electric field, it is a vector quantity. As we have two dimensions, we will have to derive in the two directions; that is

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

Performing this derivation, we get

$E(x,y)=(-300x,400y)$

For the point $(2,2)$, the electric field vector will be

$\frac{E(x,y)=(-600\hat{x}+800\widehat{y)}}{}$

The magnitude of this vector will be

$\left|\overrightarrow{E}\right|=\sqrt{{(-600)}^2+{800}^2}=1000$,

which one could also see as a multiple of $(3,4,5)$right triangle. This is to say that the magnitude of the electric field at the $(2,2)$point as$100V/m$.

The direction of the electric field will be

$a={\tan }^{-1}\frac{800}{-600}=-{53}^{\circ }$,

which means that the vector point $53$degrees lower than the direction of the positivelocalid="1648622057662" $x$axis.