Electric Field from Charge Distribution: Analyzing V(x,y,z)

[Zarrey]

Homework Statement

The electric potential for some charge distribution is given by the function V(x,y,z) = 1000z (Volts), for -10cm <= z <= +10cm, and zero otherwise (it does not depend on the x and y coordinates). Find the electric field corresponding to the given electric potential. Draw some electric field lines. What charge distribution can create such an electric field? Give all possible numerical information about the charge distribution that you can find out from the given data.

I have no clue about this one.

i know that E = σ/2ε₀. though I am not sure how it applies to this question.

 

[Sourabh N]

How are electric field (the thing you want) related to electric potential (the thing you have)? Before I give details, I'll give a short answer: Electric field is the negative gradient of electric potential. You can find the definition of gradient in case you don't know it, in your textbook/internet.

Does that help?

 

[Zarrey]

i think i found what you were referring to.

E_r = -(∂V/∂r) for the radial electric field.

is this what i should be using?

 

[Sourabh N]

i think i found what you were referring to.

E_r = -(∂V/∂r) for the radial electric field.

is this what i should be using?

Yes, with a slight modification. Your electric potential is written in cartesian coordinates, so it'll be easier to have the gradient (and the electric field) in cartesian coordinates also.

 

[Zarrey]

what do you mean by that?

 

[Sourabh N]

Near wherever you found that expression of radial electric field, should be an expression for electric field in cartesian coordinates. If not, Wikipedia goes in detail about gradients in different coordinates system. If that article is too loud, here is gradient in the three-dimensional Cartesian coordinate system:

$\nabla f$ = $\frac{df}{dx}$ i + $\frac{df}{dy}$ j + $\frac{df}{dz}$ k.

 

[Zarrey]

oh ok. i found that equation as well.

so based on the equation am i to say that the coordinates at the x and y values are some number other than 0 on the range of -10 to 10 cm for z?

am i to solve for the coordinates for x and y? if so how do i do that?

 

[Sourabh N]

so based on the equation am i to say that the coordinates at the x and y values are some number other than 0 on the range of -10 to 10 cm for z?

Which equation are you referring to?

For this question, the answer is independent of the values of x and y, so should be fine at x=0=y.

 

[Zarrey]

the equation given for the electric potential. so am i to say that the field only goes along the z axis?

 

[Zarrey]

i appreciate the time you are taking to help me by the way.

 

[Sourabh N]

Yes! If you were to draw the electric field lines, they would be equally spaced parallel lines, parallel to the z axis, pointing from negative to positive direction (or positive to negative, I'm not very sure).

And no problem!

 

[Zarrey]

the lines would be flowing from the positive source to the negative.

so the way to draw the E-field would be to set up the x,y,z coordinate system graph and draw lines that run parallel to the Z axis with arrows pointing downward because the positive is in the upper quadrant and the negative is in the lower quadrant?

how would i solve for the potential tho? am i to solve it the same way i normally would and disregard the other 2 values ( x and y ) and just use the Z value for my q value?

 

[Sourabh N]

the lines would be flowing from the positive source to the negative.

so the way to draw the E-field would be to set up the x,y,z coordinate system graph and draw lines that run parallel to the Z axis with arrows pointing downward because the positive is in the upper quadrant and the negative is in the lower quadrant?

Yes!

how would i solve for the potential tho? am i to solve it the same way i normally would and disregard the other 2 values ( x and y ) and just use the Z value for my q value?

I'm not sure what you're saying. solve for potential? q value?

 

[Zarrey]

The question is asking for numerical values for the charge distribution. i know now what the field will look like but i don't know what I am supposed to be plugging into the equation above.

It also asks to fine the electric field, does this mean solve to fine the values at each value of z or does this mean to sketch it on the graph with the lines and arrows?

 

[Sourabh N]

The question is asking for numerical values for the charge distribution. i know now what the field will look like but i don't know what I am supposed to be plugging into the equation above.

i know that E = σ/2ε₀. though I am not sure how it applies to this question.

Your intuition(?) is correct here. One of the ways to produce a constant electric field is to have two very long charged plates (or, sitting in the middle of two charged plates placed very close to each other). The charge density on each plate to produce such an electric field is given by your formula above.

It also asks to fine the electric field, does this mean solve to fine the values at each value of z or does this mean to sketch it on the graph with the lines and arrows?

Find the electric field corresponding to the given electric potential. Draw some electric field lines.

So, it means both. You already have computed the electric field, drawing a few field lines should do the job.

 

[Zarrey]

i don't have the sigma values to calculate the field though. how am i to find those?

 

[Sourabh N]

That's what you have find. You know the other two quantities in the equation.

 

[Zarrey]

ok to find the values i have to plug in different numbers into the z value for the equation

E = σ/2ε₀

E ( 2ε₀ ) = σ

1000z ( 2ε₀ ) = σ on the range of -10cm < z < +10cm or -.01m < z < .01m

 

[Sourabh N]

ok to find the values i have to plug in different numbers into the z value for the equation

E = σ/2ε₀

E ( 2ε₀ ) = σ

1000z ( 2ε₀ ) = σ on the range of -10cm < z < +10cm or -.01m < z < .01m

I think E should be -1000 <some units>?

 

[Zarrey]

why would E be negative?

 

[Sourabh N]

why would E be negative?

[...]Electric field is the negative gradient of electric potential. [...]

[...] here is gradient in the three-dimensional Cartesian coordinate system:

$\nabla f$ = $\frac{df}{dx}$ i + $\frac{df}{dy}$ j + $\frac{df}{dz}$ k.

That's why.