Question

The potential in a region of space due to a charge distribution is given by the expression V = ax2z + bxy − cz2 where a = −3.00 V/m3, b = 6.00 V/m2, and c = 9.00 V/m2. What is the electric field vector at the point (0, −8.00, −8.00) m? Express your answer in vector form

Answer 1

Answer:

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

Explanation:

We know that the relationship between the electric field $\vec{E}(\vec{r})$ and the potential $V(\vec{r})$ is given by

$\vec{E} ( \vec{r}) = - \vec{\nabla} V(\vec{r})$

So, for our potential:

$V(r) = a x^2 z + b x y - c z^2$

the electric field is :

$\vec{E} ( \vec{r}) = - \vec{\nabla} ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x}, \frac{ \partial }{\partial y} , \frac{ \partial }{\partial z}) ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial y} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial z} ( a x^2 z + b x y - c z^2 ))$

$\vec{E} ( \vec{r}) = - ( 2 a x z + b y , b x , a x^2 - 2 c z )$

This is the our electric field. At vector point

$\vec{r} = (0, -8.00 \ m, - 8.00 \ m)$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 2 a * 0 * (-8.00 \ m) + b (-8.00 \ m) , b * 0 , a 0^2 - 2 c (-8.00 \ m) )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 0 - b 8.00 \ m , 0 , 0 + 2 c 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( b 8.00 \ m , 0 , - 2 c 8.00 \ m )$

Knowing

$b= 6.00 \frac{V}{m^2}$

and

$c=9.00 \frac{V}{m^2}$

the electric field is

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 6.00 \frac{V}{m^2} * 8.00 \ m , 0 , - 9.00 \frac{V}{m^2} * 2* 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$