Question

The electrostatic potential existing in space is given by V = (xy + yz + zx) volts, where x, y and z are in metres. The angle made by electric field at (2 m, 2 m, 2 m) with positive x axis is a. Then

Answer 1

SOLUTION

GIVEN

The electrostatic potential existing in space is given by V = (xy + yz + zx) volts, where x, y and z are in metres.

TO DETERMINE

The angle made by electric field at (2 m, 2 m, 2 m) with positive x axis

EVALUATION

Here it is given that the electrostatic potential existing in space is given by V = (xy + yz + zx) volts, where x, y and z are in metres.

Now electric field

$\vec{ E} = - \nabla V$

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

$\displaystyle = - (y + z) \hat{i} - (x + z) - (x + y) \hat{k}$

Now at (2 m, 2 m, 2 m) the electric field is

$\displaystyle = - (2 + 2) \hat{i} - (2 + 2) - (2 + 2) \hat{k}$

$\displaystyle = - 4 \hat{i} - 4 \hat{j} - 4\hat{k}$

Now

$| \vec{ E}| = \sqrt{ {( - 4)}^{2} + {( - 4)}^{2} + {( - 4)}^{2} }$

$\implies | \vec{ E}| = \sqrt{ 48 }$

$\implies | \vec{ E}| = 4 \sqrt{ 3 }$

Let $\alpha$ be the required angle made by electric field at (2 m, 2 m, 2 m) with positive x axis

Then

$\displaystyle \cos \alpha = \frac{ - 4}{4 \sqrt{3} }$

$\displaystyle \implies \: \cos \alpha = - \frac{1}{ \sqrt{3} }$

$\displaystyle \implies \: \alpha = { \cos}^{ - 1} \bigg( - \frac{1}{ \sqrt{3}} \bigg)$

FINAL ANSWER

Hence the required angle

$\displaystyle = { \cos}^{ - 1} \bigg( - \frac{1}{ \sqrt{3}} \bigg)$

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Answer 2

Answer:

cos alpha=√-1/3

Explanation:

hope it will help u