If electric potential at any point is V=-6 x+2 y+9 z, where V, x, y and z are in SI units, then the magnitude of the electric field (in SI unit) is

Question

If electric potential at any point is V=-6 x+2 y+9 z, where V, x, y and z are in SI units, then the magnitude of the electric field (in SI unit) is (A) 3 √2 (B) 4 √2 (C) 5 √2 (D) 11

Tardigrade Admin · Accepted Answer

Ex=-(d V/d x)=-(-6)=6 ; Ey=-(d V/d y)=-2 and Ez=-(d V/d z)=-9 E text net =√Ex2+Ey2+Ez2 =√36+4+81=√121=11

Q. If electric potential at any point is $V = - 6 x + 2 y + 9 z$ , where $V, x, y$ and $z$ are in SI units, then the magnitude of the electric field (in SI unit) is

$32$

$42$

$52$

11

$E_{x} = - \frac{d V}{d x} = - (- 6) = 6;$
$E_{y} = - \frac{d V}{d y} = - 2$
and $E_{z} = - \frac{d V}{d z} = - 9$
$E_{net} = E_{x}^{2} + E_{y}^{2} + E_{z}^{2}$
$= 36 + 4 + 81 = 121 = 11$

Q. If electric potential at any point is V=−6x+2y+9z, where V,x,y and z are in SI units, then the magnitude of the electric field (in SI unit) is

32​

42​

52​

11

Ex​=−dxdV​=−(−6)=6; Ey​=−dydV​=−2 and Ez​=−dzdV​=−9 Enet ​=Ex2​+Ey2​+Ez2​​ =36+4+81​=121​=11

Q. If electric potential at any point is $V = - 6 x + 2 y + 9 z$ , where $V, x, y$ and $z$ are in SI units, then the magnitude of the electric field (in SI unit) is

$32$

$42$

$52$

$E_{x} = - \frac{d V}{d x} = - (- 6) = 6;$
$E_{y} = - \frac{d V}{d y} = - 2$
and $E_{z} = - \frac{d V}{d z} = - 9$
$E_{net} = E_{x}^{2} + E_{y}^{2} + E_{z}^{2}$
$= 36 + 4 + 81 = 121 = 11$