Question #a9813

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Answer 1 · 2016-05-10T07:34:06

$E = E_{x} = - 12 \frac{V}{m}$ $E=E_x=-12 V/m$

$given: V = 3 x^{2}$ $"given: "V=3x^2$

$E_{x} = - \frac{d V}{d x}$ $E_x=-(d V)/(d x)$

$E_{y} = 0$ $E_y=0$

$E_{z} = 0$ $E_z=0$

$E_{x} = - \frac{d (3 x^{2})}{d x}$ $E_x=-(d(3x^2))/(d x)$

$E_{x} = - 6 x$ $E_x=-6x$

$P=(color(green)(2),cancel(0),cancel(1))$

$E_x=-6*2$

$E_x=-12$

$E=E_x=-12 V/m$

Answer 2 · 2016-05-12T13:54:20

$-12hatiV/m$

Potentoal at any point (x,y,z) ls
$V(x,y,z) =3x^2$
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 $( deltaV)/(deltay)=0 and ( deltaV)/(deltaz)=0$

$vecE=-hati(deltaV)/(deltax)$
$vecE=-(hatidelta(3x^2))/(deltax)$

$vecE=-hati(6x)$
At the given point $(2m,0,1m)$
the Electric field is(for x=2)

$vecE=-hati(6xx2)=-12hati V/m$
,if the umit of potential is Volt