HELP A problem on determine electric field in vector form

In summary: Your derivatives are wrong. If you post the steps you took to calculate them then I can show you where you made a mistake.
  • #1
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
 
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  • #2


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 [itex]V(x,y)[/itex] at the given point. And even if you were supposed to do that, it makes no sense to insert the unit vectors [itex]\hat{i}[/itex] and [itex]\hat{j}[/itex] 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.
 
  • #3


Tom Mattson said:
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 [itex]V(x,y)[/itex] at the given point. And even if you were supposed to do that, it makes no sense to insert the unit vectors [itex]\hat{i}[/itex] and [itex]\hat{j}[/itex] 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
Ex=-[tex]\delta[/tex]V/[tex]\delta[/tex]x
Ex=-(100/(3/2(x2+y2)1/2*(2x+y2))
Ey=-[tex]\delta[/tex]V/[tex]\delta[/tex]y
Ey=(100x/(3/2(x2+y2)1/2*(x2+2y))
then sub in x and y
is it right?
 
  • #4


Your derivatives are wrong. If you post the steps you took to calculate them then I can show you where you made a mistake.
 
  • #5


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

Ex=-[tex]\delta[/tex]V/[tex]\delta[/tex]x
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)
Ex=-(100/(3/2(x2+y2)1/2*(2x))
same for y but for this only derive y and let x be constant
Ey=-[tex]\delta[/tex]V/[tex]\delta[/tex]y
Ey=(100x/(3/2(x2+y2)1/2*(2y))
 

1. What is the equation for determining electric field in vector form?

The equation for determining electric field in vector form is E = F/q, where E is the electric field, F is the force, and q is the charge.

2. How do you find the direction of the electric field using vector form?

To find the direction of the electric field using vector form, you must consider the direction of the force acting on the charge. The electric field will point in the same direction as the force acting on the charge.

3. What is the unit of measurement for electric field in vector form?

The unit of measurement for electric field in vector form is newtons per coulomb (N/C).

4. Can electric field be negative in vector form?

Yes, electric field can be negative in vector form. This indicates that the direction of the electric field is opposite to the direction of the force acting on the charge.

5. How does distance affect the strength of the electric field in vector form?

The strength of the electric field in vector form is inversely proportional to the square of the distance from the charge. This means that as the distance increases, the electric field becomes weaker.

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