If electric potential at any point is V=-5 x+3 y+√15 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=-5 x+3 y+√15 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) 7

Tardigrade Admin · Accepted Answer

Ex=-(d V/d x)=-(-5)=5 ; Ey=-(d V/d y)=-3 and Ez=-(d V/d z)=-√15 E text net =√Ex2+Ey2+Ez2 =√(5)2+(-3)2+(-√15)2=7

Q. If electric potential at any point is $V = - 5 x + 3 y + 15 z$ , where $V, x, y$ and $z$ are in SI units, then the magnitude of the electric field (in SI unit) is

$32$

$42$

$52$

7

$E_{x} = - \frac{d V}{d x} = - (- 5) = 5;$
$E_{y} = - \frac{d V}{d y} = - 3$
and $E_{z} = - \frac{d V}{d z} = - 15$
$E_{net} = E_{x}^{2} + E_{y}^{2} + E_{z}^{2}$
$= (5)^{2} + (- 3)^{2} + (- 15)^{2} = 7$

Q. If electric potential at any point is V=−5x+3y+15​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​

7

Ex​=−dxdV​=−(−5)=5; Ey​=−dydV​=−3 and Ez​=−dzdV​=−15​ Enet ​=Ex2​+Ey2​+Ez2​​ =(5)2+(−3)2+(−15​)2​=7

Q. If electric potential at any point is $V = - 5 x + 3 y + 15 z$ , where $V, x, y$ and $z$ are in SI units, then the magnitude of the electric field (in SI unit) is

$32$

$42$

$52$

$E_{x} = - \frac{d V}{d x} = - (- 5) = 5;$
$E_{y} = - \frac{d V}{d y} = - 3$
and $E_{z} = - \frac{d V}{d z} = - 15$
$E_{net} = E_{x}^{2} + E_{y}^{2} + E_{z}^{2}$
$= (5)^{2} + (- 3)^{2} + (- 15)^{2} = 7$