HELP A problem on determine electric field in vector form

[Kudo Shinichi]

HELP!A problem on determine electric field in vector form

Homework Statement

The electric potential due to a charge distribution is given by V(x,y)=((100x)/(x²+y²)^3/2)*Volts
where the distances are in meters. What is the electric field (vector) at the position x=2m, y=0.2m?

The Attempt at a Solution

V(x,y)=((100*2i)/(2²i+0.2²j)^3/2)*volts
V(x,y)=(200i/(8i+0.008j))*volts
This is how I tried to solve the problem but I don't think that I did it correctly because I don't think that I can write a function with two directions together.
Can anyone help me with it? thank you very much

 

[quantumdude]

One thing is clear: You aren't reading your textbook.

First, they ask for the electric field, not the electric potential. So you are not supposed to evaluate V(x,y) at the given point. And even if you were supposed to do that, it makes no sense to insert the unit vectors \hat{i} and \hat{j} in there (where would those come from anyway??).

Please look in your book and find the definition of electric field in terms of electric potential. Then we can get started.

 

[Kudo Shinichi]

One thing is clear: You aren't reading your textbook.

First, they ask for the electric field, not the electric potential. So you are not supposed to evaluate V(x,y) at the given point. And even if you were supposed to do that, it makes no sense to insert the unit vectors \hat{i} and \hat{j} in there (where would those come from anyway??).

Please look in your book and find the definition of electric field in terms of electric potential. Then we can get started.

Sorry about that I didn't read the problem properly...
dV=-E*dl
E=-dV/dl
Then separate the x and y direction
E_x=-\deltaV/\deltax
E_x=-(100/(3/2(x²+y²)^1/2*(2x+y²))
E_y=-\deltaV/\deltay
E_y=(100x/(3/2(x²+y²)^1/2*(x²+2y))
then sub in x and y
is it right?

 

[quantumdude]

Your derivatives are wrong. If you post the steps you took to calculate them then I can show you where you made a mistake.

 

[Kudo Shinichi]

Your derivatives are wrong. If you post the steps you took to calculate them then I can show you where you made a mistake.

E_x=-\deltaV/\deltax
For this only we only have to derive x and let y be constant
the original equation is 100x/(x²+y²)^3/2
since we only need to derive x
then the equation becomes
100x-->100
use chain rule to do the following=outside derivative times inside derivative
(x²+y²)^3/2
3/2(x²+y²)^1/2*(2x)
E_x=-(100/(3/2(x²+y²)^1/2*(2x))
same for y but for this only derive y and let x be constant
E_y=-\deltaV/\deltay
E_y=(100x/(3/2(x²+y²)^1/2*(2y))