Question

*＊✿❀Subh Sandhya mitron.....❀✿＊*



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Answer 1

Given:

$\large\rm { x = -1m, y = 2m, z = 3m}$

Solution:

we know that $\large\rm { E_{x} = \frac{ - \partial V}{ \partial x} }$

substituting the values

$\large\rm { \frac{- \partial}{\partial x} (10x^{2} + 5y^{2} - 3z^{2})}$

$\large\rm { = -20 x = -20(-1) \boxed{ = \pm 20 Vm^{-1}}}$

$\large\rm { \ }$

$\large\rm { \ }$

now

$\large\rm { E_{y} = \frac{ - \partial V}{ \partial y} }$

substituting the values

$\large\rm { \frac{- \partial}{\partial y} (10x^{2} + 5y^{2} - 3z^{2})}$

$\large\rm{ -10(y) = -10(2) \boxed{= -20 Vm^{-1}}}$

$\large\rm { \ }$

$\large\rm { \ }$

now

$\large\rm { E_{z} = \frac{ - \partial V}{ \partial z} }$

substituting the values

$\large\rm { \frac{- \partial}{\partial z} (10x^{2} + 5y^{2} - 3z^{2})}$

$\large\rm { = 6(z) = 6(3) \boxed{ = 18 Vm^{-1}}}$

Answer 2

hope this helps you

hello Vishal