Question a9813

May 10, 2016

$E = {E}_{x} = - 12 \frac{V}{m}$

Explanation:

$\text{given: } V = 3 {x}^{2}$

${E}_{x} = - \frac{d V}{d x}$

${E}_{y} = 0$

${E}_{z} = 0$

${E}_{x} = - \frac{d \left(3 {x}^{2}\right)}{d x}$

${E}_{x} = - 6 x$

$P = \left(\textcolor{g r e e n}{2} , \cancel{0} , \cancel{1}\right)$

${E}_{x} = - 6 \cdot 2$

${E}_{x} = - 12$

$E = {E}_{x} = - 12 \frac{V}{m}$

May 12, 2016

$- 12 \hat{i} \frac{V}{m}$

Explanation:

Potentoal at any point (x,y,z) ls
$V \left(x , y , z\right) = 3 {x}^{2}$
We know
vecE=-(hati(deltaV)/(deltax)+hatj(deltaV)/(deltay)+hatk( deltaV)/(deltaz))#

The given function of Potential ‘V’ is independent of y and z.
So $\frac{\delta V}{\delta y} = 0 \mathmr{and} \frac{\delta V}{\delta z} = 0$

$\vec{E} = - \hat{i} \frac{\delta V}{\delta x}$
$\vec{E} = - \frac{\hat{i} \delta \left(3 {x}^{2}\right)}{\delta x}$

$\vec{E} = - \hat{i} \left(6 x\right)$
At the given point $\left(2 m , 0 , 1 m\right)$
the Electric field is(for x=2)

$\vec{E} = - \hat{i} \left(6 \times 2\right) = - 12 \hat{i} \frac{V}{m}$
,if the umit of potential is Volt