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Lady_Fox [76]
2 years ago
14

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
Physics
1 answer:
Elis [28]2 years ago
7 0

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}  )

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