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

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

Step-by-step 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{\\abla} 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{\\abla} ( a x^2 z + b x y - c z^2 )`

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

`\vec{E} ( \vec{r}) = - ( ( \partial )/(\partial x) ( a x^2 z + b x y - c z^2 ) , ( \partial )/(\partial y)  ( a x^2 z + b x y - c z^2 ) , ( \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 (V)/(m^2)`

and

`c=9.00 (V)/(m^2)`

the electric field is

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

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