Question

Electric potential in a particular region of space is V=12x-3x^(2)y+2yz^(2) .The electric field at point P(1m 0 -2m) is : 1) 12 unit 213 unit 35 unit 4) zero​

Answer 1

Given:

V = 12x - 3x²y + 2yz²

To find:

Electrostatic field intensity at (1, 0, 2) metres?

Calculation:

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

$\implies \: E_{x} = - \dfrac{ \partial (12x - 3 {x}^{2}y + 2 {y}^{2}z) }{ \partial x}$

$\implies\: E_{x} = - 12 + 6xy + 0$

$\implies\: E_{x} \bigg|_{x = 1} = - 12 + 6(1)(0)+ 0$

$\implies\: E_{x} = - 12 \: N/C$

_______________

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

$\implies \: E_{y} = - \dfrac{ \partial (12x - 3 {x}^{2}y + 2 {y}^{2}z) }{ \partial y}$

$\implies \: E_{y} =3 {x}^{2} - 2 {z}^{2}$

$\implies \: E_{y} =3 {(1)}^{2} - 2 {(2)}^{2}$

$\implies \: E_{y} =3 - 8$

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

_______________

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

$\implies \: E_{z} = - \dfrac{ \partial (12x - 3 {x}^{2}y + 2 {y}^{2}z) }{ \partial z}$

$\implies \: E_{z} = - 2 {y}^{2}$

$\implies \: E_{z} = - 2 {(0)}^{2}$

$\implies \: E_{z} = 0 \: N/C$

Net electrostatic field intensity:

$\therefore \: E_{net} = \sqrt{ {( - 12)}^{2} + {( - 5)}^{2} + {(0)}^{2} }$

$\implies \: E_{net} = \sqrt{ {( - 12)}^{2} + {( - 5)}^{2}}$

$\implies \: E_{net} = \sqrt{169}$

$\implies \: E_{net} = 13 \: N/C$

So, net field intensity is 13 N/C.