Question

If the potential is V(x) = 2xy2, then the Y-component of electric Field is: Select one: a. -4xy b. 4y c. -2xy2 d. +4xy e. -2y2

Answer 1

Given:

The potential is V(x) = 2xy².

To find:

Y component of Electrostatic Field Intensity.

Calculation:

Y component of Electrostatic Field Intensity can be calculated by finding out the partial derivative of the Electrostatic Potential with respect to y ;

$\therefore \: V = 2x {y}^{2}$

$= > \vec{ E} = - \dfrac{ \partial V}{ \partial y} \: \hat{j}$

$= > \vec{ E} = - \dfrac{ \partial (2x {y}^{2}) }{ \partial y} \: \hat{j}$

Performing partial derivative ;

$= > \vec{ E} = - (4xy)\: \hat{j}$

So, final answer is :

$\boxed{ \sf{ \large{ \red{\vec{ E} = - (4xy)\: \hat{j}}}}}$

Note:

• While performing partial derivative w.r.t a variable , consider the other variables to be constant .

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

If V is electric potential then Electric Field is given by

$\displaystyle \: \sf{ \: \vec{E} \: = - \bigg( \frac{ \partial V\: }{ \partial x} \hat{i} + \frac{ \partial V\: }{ \partial y} \hat{j} +\frac{ \partial V\: }{ \partial z} \hat{k} \: \bigg) }$

GIVEN

If the potential is

$\sf{V(x) = 2x {y}^{2} }$

TO DETERMINE

The electric field

CALCULATION

Here

$\sf{V(x) = 2x {y}^{2} }$

Now

$\displaystyle \: \sf{ \: \frac{ \partial V\: }{ \partial x} = 2 {y}^{2} }$

$\displaystyle \: \sf{ \: \frac{ \partial V\: }{ \partial y} = 4xy}$

$\displaystyle \: \sf{ \: \frac{ \partial V\: }{ \partial z} = 0}$

Hence the electric field is

$\displaystyle \: \sf{ \: \vec{E} \: = - \bigg(2 {y}^{2} \: \hat{i} + 4xy \: \hat{j} +0 \: \hat{k} \: \bigg) }$

RESULT

$\boxed{ \sf{ \: Y - component \: of \: electric \: Field \: is \: = - \: 4xy \: \hat{j}\: }}$