Question

if the electric potential function at a point is given as,V=(-3x+4y+5z)V,then the magnitude of electric field is ______if the electric potential function at a point is given by V=(-3x+4y+5z)is represented by V,then the magnitude of the electric field is (a) 16V/m (b) 6V/M (c) 5root2 V/m (d) 5/root2 V/m please help me!!

Answer 1

Given:

the electric potential function at a point is given as,V=(-3x+4y+5z)V.

To find:

Magnitude of Electrostatic Field intensity?

Calculation:

Field intensity along X axis :

$\therefore \: E_{x} = - \dfrac{ \partial V}{ \partial x}$

$\implies \: E_{x} = - \dfrac{ \partial ( - 3x + 4y + 5z)}{ \partial x}$

$\implies \: E_{x} = - ( - 3)$

$\implies \: E_{x} = 3 \: N/C$

Field Intensity along Y axis:

$\therefore \: E_{y} = - \dfrac{ \partial V}{ \partial y}$

$\implies \: E_{y} = - \dfrac{ \partial ( - 3x + 4y + 5z)}{ \partial y}$

$\implies \: E_{x} = -4 \: N/C$

Field Intensity along Z axis:

$\therefore \: E_{z} = - \dfrac{ \partial V}{ \partial z}$

$\implies \: E_{z} = - \dfrac{ \partial ( - 3x + 4y + 5z)}{ \partial z}$

$\implies \: E_{z} = -5 \: N/C$

So, net field Intensity:

$\therefore \: E_{net} = \sqrt{ {(E_{x})}^{2} + {(E_{y})}^{2} + {(E_{z})}^{2} }$

$\implies \: E_{net} = \sqrt{ {( 3)}^{2} + {( - 4)}^{2} + {( - 5)}^{2} }$

$\implies \: E_{net} = \sqrt{ 50 }$

$\implies \: E_{net} = 5 \sqrt{2} \: N/C$

So, net Field Intensity is 5√2 N/C.